CBSE Class 10 Mathematics Real Numbers Worksheet Set 03

Question. The sum of exponents of prime factors in the prime-factorisation of 196 is:
(a) 3
(b) 4
(c) 5
(d) 6
Answer : B

Question. The HCF and the LCM of 12, 21, 15 respectively are
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Answer : C

Question. 7 × 11 × 13 × 15 + 15 is a:
(a) Composite number
(b) Whole number
(c) Prime number
(d) (a) and (b) both
Answer : D

Question. LCM of (2³ × 3 × 5) and (2⁴ × 5 × 7) is
(a) 40
(b) 560
(c) 1120
(d) 1680
Answer : D

Question. If two positive integers a and b are written as a = x³y² and b = xy³, where x and y are prime numbers, then the HCF (a, b) is:
(a) xy
(b) xy²
(c) x³y³
(d) x²y²
Answer : B

Question. If two positive integers p and q can be expressed as p = ab² and q = a³b where a and b are prime numbers, then the LCM (p, q) is:
(a) ab
(b) a²b²
(c) a³b²
(d) a³b³
Answer : C

Question. The decimal representation of 11/2³ × 5 will:
(a) terminate after 1 decimal place
(b) terminate after 2 decimal places
(c) terminate after 3 decimal places
(d) not terminate
Answer : C

Question. The total number of factors of a prime number is
(a) 1
(b) 0
(c) 2
(d) 3
Answer : C

Question. The LCM of smallest two digit composite number and smallest composite number is:
(a) 12
(b) 4
(c) 20
(d) 44
Answer : C

Question. The cube of any positive integer is not of the form:
(a) 9q
(b) 9q + 1
(c) 9q + 3
(d) 9q + 8
Answer : C

Question. If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
(a) 1
(b) 2
(c) 3
(d) 4
Answer : D

Question. The product of a non–zero rational and an irrational number is:
(a) always irrational
(b) always rational
(c) rational or irrational
(d) one
Answer : A

Question. 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75, what is the HCF of (525, 3000)?
(a) 25
(b) 125
(c) 75
(d) 15
Answer : C

Question. The decimal expansion of the rational number 14587/1250 will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Answer : D

Question. 1.23451326... is
(a) an integer
(b) an irrational number
(c) a rational number
(d) none of these
Answer : B

Question. The number of decimal places after which the decimal expansion of the rational number 9/2⁴ × 5 will terminate, is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer : D

Question. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is:
(a) 10
(b) 100
(c) 504
(d) 2520
Answer : D

Question. If HCF (a, b) = 45 and a × b = 30375, then LCM (a, b) is:
(a) 1875
(b) 1350
(c) 625
(d) 675
Answer : D

Question. If HCF of two numbers is 1, the numbers are called relatively ......... or ......... .
(a) Prime, co-prime
(b) Composite, prime
(c) Both (a) and (b)
(d) None of the above
Answer : A

Case Study Based Questions

I. The Army Day is celebrated on 15th January every year in India. The day is celebrated in the form of parades and other military shows in the national capital New Delhi as well as in all headquarters of army.

Parade I: An Army contingent of 616 members is to march behind an army band of 32 members in parade. The two groups are to march in the same number of columns.
Parade II: An Army contingent of 1000 members is to march behind an army band of 56 members in parade. The two groups are to march in the same number of columns.

Refer to Parade I

Question. Number 616 can be expressed as a product of its prime factors as
(a) 2¹ × 14¹ × 22¹
(b) 2² × 11¹ × 14¹
(c) 2³ × 7¹ × 11¹
(d) 2⁴ × 7² × 11¹
Answer : C

Question. The HCF of 32 and 616 is
(a) 8
(b) 16
(c) 18
(d) 12
Answer : A

Refer to Parade II

Question. The LCM of 56 and 1000 is
(a) 6000
(b) 7000
(c) 8000
(d) 9000
Answer : B

Question. Number 1000 can be expressed as a product of its prime factors as
(a) 2³ × 5³
(b) 2² × 5⁴
(c) 2⁴ × 5²
(d) 2³ × 5⁴
Answer : A

Question. The maximum number of columns in which army can march is
(a) 6
(b) 10
(c) 12
(d) 8
Answer : D

II. Traffic Lights are used to control movement of traffics. They are installed at crossings and intersections of roads. Usually three different colours of lights (Red, Yellow and Green) are used to tell commuters what to do.
The traffic lights at different road crossings change after every 48 seconds, 72 seconds and 120 seconds respectively.

Question. 120 can be expressed as a product of its prime factors as
(a) 2³ × 3¹ × 5¹
(b) 2² × 3² × 5¹
(c) 2² × 3¹ × 5²
(d) 2³ × 3² × 5¹
Answer : A

Question. The HCF of 48, 72 and 120 is
(a) 12
(b) 16
(c) 18
(d) 24
Answer : D

Question. The LCM of 48, 72 and 120 is
(a) 432
(b) 420
(c) 720
(d) 840
Answer : C

Question. If all the traffic lights change simultaneously at 7 : 30 : 00 hours, they will change again simultaneously at
(a) 7 : 42 : 00 hrs
(b) 7 : 52 : 00 hrs
(c) 7 : 36 : 00 hrs
(d) 7 : 46 : 00 hrs
Answer : A

Question. The [HCF × LCM] for the numbers 48, 72 and 120 is
(a) 17480
(b) 17280
(c) 12280
(d) 18280
Answer : B

1. If 7x5x3x2 + 3 is composite number? Justify your answer

2. Show that any positive odd integer is of the form 4q + 1 or 4q +3 where q is a positive integer

3. Prove that √2 + √5 is irrational

4. Prove that 5 - 2√3 is an irrational number

5. Prove that √2 is irrational

6. Use Euclid’s Division Algorithms to find the H.C.F of a) 135 and 225 (45)

b) 4052 and 12576 (4)

c) 270, 405 and 315 (45)

7. Find the HCF and LCM of 26 and 91 and verify that LCM X HCF = Product of two numbers (13,182) 8. Explain why 29 is a terminating decimal expansion 23 x 53 

9. 163 will have a terminating decimal expansion. State true or false .Justify your answer. 150

10. Find HCF of 96 and 404 by prime factorization method. Hence, find their LCM. (4, 9696)

11. Using prime factorization method find the HCF and LCM of 72, 126 and 168 (6, 504)

12. If HCF (6, a) = 2 and LCM (6, a) = 60 then find a (20)

13. given that LCM (77, 99) = 693, find the HCF (77, 99) (11)

14. Find the greatest number which exactly divides 280 and 1245 leaving remainder 4 and 3 (138)

15. The LCM of two numbers is 64699, their HCF is 97 and one of the numbers is 2231. Find the other (2813)

16. Two numbers are in the ratio 15: 11. If their HCF is 13 and LCM is 2145 then find the numbers (195,143)

17. Express 0.363636………… in the form a/b (4/11)

18. Write the HCF of smallest composite number and smallest prime number

19. Write whether 2√45 + 3√20 on simplification give a rational or an irrational number 2√5

20. State whether 10.064 is rational or not. If rational, express in p/q form

21. Write a rational number between √2 and √3

22. State the fundamental theorem of arithmetic PREPARED BY: MAHABOOB PASHA IX – X BOYS

Question 1. If 7x5x3x2 + 3 is composite number? Justify your answer
Answer: Let us simplify the given expression by factoring out 3:
\( 7 \times 5 \times 3 \times 2 + 3 = 3 \times (7 \times 5 \times 2 + 1) \)
\( \implies 3 \times (70 + 1) \)
\( \implies 3 \times 71 \)
Since the number has prime factors other than 1 and itself (3 and 71), it is a composite number.
In simple words: We can pull out 3 as a common factor, making the number a product of 3 and 71. Since it has factors other than 1 and itself, it is composite.

Exam Tip: Always factor out the common term to express the number as a product of two or more prime factors when justifying composite numbers.

Question 2. Show that any positive odd integer is of the form 4q + 1 or 4q +3 where q is a positive integer
Answer: According to Euclid's division lemma, let \( a \) be any positive odd integer and let \( b = 4 \).
Then we can write:
\( a = 4q + r \), where \( 0 \le r < 4 \) and \( q \ge 0 \) is an integer.
The possible remainders are \( r = 0, 1, 2, 3 \).
Let us analyze each case:
- If \( r = 0 \), then \( a = 4q \), which is divisible by 2 and is therefore an even integer.
- If \( r = 1 \), then \( a = 4q + 1 \), which is not divisible by 2 and is therefore an odd integer.
- If \( r = 2 \), then \( a = 4q + 2 = 2(2q + 1) \), which is divisible by 2 and is therefore an even integer.
- If \( r = 3 \), then \( a = 4q + 3 \), which is not divisible by 2 and is therefore an odd integer.
Since \( a \) is a positive odd integer, it cannot be of the form \( 4q \) or \( 4q + 2 \).
Thus, any positive odd integer is of the form \( 4q + 1 \) or \( 4q + 3 \).
In simple words: When we divide any odd number by 4, the remainder must be either 1 or 3 because even remainders would make the entire number even.

Exam Tip: Clearly state Euclid's division lemma formula and specify the range of the remainder \( r \) to secure step-wise marks.

Question 3. Prove that √2 + √5 is irrational
Answer: Let us assume on the contrary that \( \sqrt{2} + \sqrt{5} \) is a rational number.
Let \( \sqrt{2} + \sqrt{5} = a \), where \( a \) is a rational number.
Squaring both sides of the equation:
\( (\sqrt{2} + \sqrt{5})^2 = a^2 \)
\( \implies 2 + 5 + 2\sqrt{10} = a^2 \)
\( \implies 7 + 2\sqrt{10} = a^2 \)
\( \implies 2\sqrt{10} = a^2 - 7 \)
\( \implies \sqrt{10} = \frac{a^2 - 7}{2} \)
Since \( a \) is a rational number, \( \frac{a^2 - 7}{2} \) must also be a rational number.
This implies that \( \sqrt{10} \) is rational, which is a contradiction because the square root of a non-perfect square integer is always irrational.
Therefore, our assumption is incorrect, and \( \sqrt{2} + \sqrt{5} \) is irrational.
In simple words: If we assume the sum is rational, squaring both sides leads to an equation stating that the irrational number \( \sqrt{10} \) is rational, which is impossible.

Exam Tip: Be sure to write the contradiction statement clearly, explaining why a rational term cannot equal an irrational term.

Question 4. Prove that 5 - 2√3 is an irrational number
Answer: Let us assume on the contrary that \( 5 - 2\sqrt{3} \) is a rational number.
We can write it in the form \( \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers with \( b \neq 0 \).
\( \implies 5 - 2\sqrt{3} = \frac{a}{b} \)
\( \implies 2\sqrt{3} = 5 - \frac{a}{b} \)
\( \implies 2\sqrt{3} = \frac{5b - a}{b} \)
\( \implies \sqrt{3} = \frac{5b - a}{2b} \)
Since \( a \) and \( b \) are integers, the term \( \frac{5b - a}{2b} \) is rational.
This implies that \( \sqrt{3} \) is rational, which contradicts the fact that \( \sqrt{3} \) is irrational.
Hence, our assumption is false, and \( 5 - 2\sqrt{3} \) is an irrational number.
In simple words: Rearranging the expression lets us isolate the irrational part \( \sqrt{3} \) on one side, showing that it equals a rational fraction, which is a contradiction.

Exam Tip: Isolate the irrational radical on one side of the equation and the rational terms on the other side to complete the contradiction proof smoothly.

Question 5. Prove that √2 is irrational
Answer: Let us assume on the contrary that \( \sqrt{2} \) is rational.
Then we can write:
\( \sqrt{2} = \frac{a}{b} \), where \( a \) and \( b \) are co-prime integers and \( b \neq 0 \).
Squaring both sides of the equation:
\( 2 = \frac{a^2}{b^2} \implies a^2 = 2b^2 \)
This shows that \( a^2 \) is divisible by 2, which implies that \( a \) is also divisible by 2.
Let \( a = 2c \) for some integer \( c \).
Substituting this back into the equation:
\( (2c)^2 = 2b^2 \)
\( \implies 4c^2 = 2b^2 \)
\( \implies b^2 = 2c^2 \)
This shows that \( b^2 \) is divisible by 2, which means that \( b \) must also be divisible by 2.
Since both \( a \) and \( b \) have a common factor of 2, they are not co-prime.
This contradicts our initial assumption that \( a \) and \( b \) are co-prime.
Thus, \( \sqrt{2} \) is irrational.
In simple words: Assuming \( \sqrt{2} \) is rational leads to the conclusion that both the numerator and denominator share a common factor of 2, contradicting our rule that the fraction must be in its simplest form.

Exam Tip: Do not miss the step of showing both variables are divisible by 2, as this is the core argument of the proof by contradiction.

Question 6. Use Euclid’s Division Algorithms to find the H.C.F of a) 135 and 225 b) 4052 and 12576 c) 270, 405 and 315
Answer:
a) 135 and 225:
Using Euclid's Division Lemma:
\( 225 = 135 \times 1 + 90 \)
\( 135 = 90 \times 1 + 45 \)
\( 90 = 45 \times 2 + 0 \)
Since the remainder is 0, the divisor at this step is the HCF.
\( \implies \text{HCF}(135, 225) = 45 \)

b) 4052 and 12576:
Using Euclid's Division Lemma:
\( 12576 = 4052 \times 3 + 420 \)
\( 4052 = 420 \times 9 + 272 \)
\( 420 = 272 \times 1 + 148 \)
\( 272 = 148 \times 1 + 124 \)
\( 148 = 124 \times 1 + 24 \)
\( 124 = 24 \times 5 + 4 \)
\( 24 = 4 \times 6 + 0 \)
Since the remainder is 0, the final divisor is the HCF.
\( \implies \text{HCF}(4052, 12576) = 4 \)

c) 270, 405 and 315:
First, let us find the HCF of 270 and 315 using Euclid's lemma:
\( 315 = 270 \times 1 + 45 \)
\( 270 = 45 \times 6 + 0 \)
Thus, the HCF of 270 and 315 is 45.
Now, let us find the HCF of 45 and 405:
\( 405 = 45 \times 9 + 0 \)
Thus, the HCF of all three numbers is 45.
In simple words: We divide the larger number by the smaller one repeatedly, taking the remainder as the new divisor, until we get a remainder of 0.

Exam Tip: Write down every single step of the division process. Leaving out steps can result in loss of working marks.

Question 7. Find the HCF and LCM of 26 and 91 and verify that LCM X HCF = Product of two numbers (13,182)
Answer: Let us write the prime factorizations:
\( 26 = 2 \times 13 \)
\( 91 = 7 \times 13 \)
HCF is the product of common prime factors with the lowest power:
\( \implies \text{HCF} = 13 \)
LCM is the product of prime factors with the highest power:
\( \implies \text{LCM} = 2 \times 7 \times 13 = 182 \)
Now let us verify the formula:
\( \text{LCM} \times \text{HCF} = 182 \times 13 = 2366 \)
\( \text{Product of numbers} = 26 \times 91 = 2366 \)
Since \( 2366 = 2366 \), the relation is verified.
In simple words: The common factor is 13, and the combined multiple is 182. Multiplying them together gives the same result as multiplying the original two numbers.

Exam Tip: For verification questions, clearly display LHS and RHS evaluations separately to show they are equal.

Question 8. Explain why 29 is a terminating decimal expansion / 2³ x 5³
Answer: Let us write the given fraction:
\( \frac{29}{2^3 \times 5^3} \)
A rational number in its simplest form \( \frac{p}{q} \) has a terminating decimal expansion if the prime factorization of its denominator \( q \) is of the form \( 2^n \times 5^m \), where \( n \) and \( m \) are non-negative integers.
Here, the denominator \( 2^3 \times 5^3 \) contains only powers of 2 and 5.
Therefore, the rational number has a terminating decimal expansion.
In simple words: Since the denominator consists only of factors of 2 and 5, dividing 29 by this denominator will result in a decimal that ends.

Exam Tip: State the rule that the denominator must be of the form \( 2^n \times 5^m \) to make your explanation complete and theoretically sound.

Question 9. 163 will have a terminating decimal expansion. State true or false .Justify your answer. / 150
Answer: The given rational number is:
\( \frac{163}{150} \)
Let us find the prime factorization of the denominator:
\( 150 = 2 \times 3 \times 5^2 \)
Since the denominator contains a prime factor other than 2 and 5 (which is 3), the rational number will have a non-terminating repeating decimal expansion.
Thus, the given statement is False.
In simple words: Since 150 has 3 as a prime factor, it does not fit the requirement of having only 2s and 5s in its denominator, making the decimal non-terminating.

Exam Tip: Always factor the denominator completely to check if any prime factors other than 2 and 5 exist.

Question 10. Find HCF of 96 and 404 by prime factorization method. Hence, find their LCM. (4, 9696)
Answer: Let us write the prime factorizations:
\( 96 = 2^5 \times 3 \)
\( 404 = 2^2 \times 101 \)
\( \implies \text{HCF} = 2^2 = 4 \)
Using the relationship between HCF and LCM:
\( \text{LCM} \times \text{HCF} = \text{Product of the two numbers} \)
\( \implies \text{LCM} \times 4 = 96 \times 404 \)
\( \implies \text{LCM} = \frac{96 \times 404}{4} = 96 \times 101 = 9696 \)
Thus, the HCF is 4 and LCM is 9696.
In simple words: Prime factors show that the highest common factor is 4. We can use this to divide the product of the numbers to easily find the LCM as 9696.

Exam Tip: If the question asks to "Hence find their LCM", you must use the product formula rather than starting the prime factorization method again for the LCM.

Question 11. Using prime factorization method find the HCF and LCM of 72, 126 and 168 (6, 504)
Answer: Let us find the prime factorizations:
\( 72 = 2^3 \times 3^2 \)
\( 126 = 2 \times 3^2 \times 7 \)
\( 168 = 2^3 \times 3 \times 7 \)
HCF is the product of common prime factors with the lowest power:
\( \implies \text{HCF} = 2^1 \times 3^1 = 6 \)
LCM is the product of all prime factors with their highest power:
\( \implies \text{LCM} = 2^3 \times 3^2 \times 7 = 8 \times 9 \times 7 = 504 \)
Thus, the HCF is 6 and the LCM is 504.
In simple words: By finding the prime factors of all three numbers, we get the common factor 6 and the first common multiple 504.

Exam Tip: Write down all prime factors clearly in exponent form to make identifying lowest and highest powers straightforward.

Question 12. If HCF (6, a) = 2 and LCM (6, a) = 60 then find a (20)
Answer: We use the formula linking two numbers to their HCF and LCM:
\( \text{HCF}(6, a) \times \text{LCM}(6, a) = 6 \times a \)
\( \implies 2 \times 60 = 6 \times a \)
\( \implies 120 = 6a \)
\( \implies a = 20 \)
Thus, the value of \( a \) is 20.
In simple words: Since the product of two numbers is equal to the product of their HCF and LCM, we divide 120 by 6 to find the missing number, which is 20.

Exam Tip: This equation is very straightforward, but double-check your division step to make sure you do not make a basic math error.

Question 13. given that LCM (77, 99) = 693, find the HCF (77, 99) (11)
Answer: We use the relation:
\( \text{HCF} \times \text{LCM} = \text{Product of the two numbers} \)
\( \implies \text{HCF} \times 693 = 77 \times 99 \)
\( \implies \text{HCF} = \frac{77 \times 99}{693} \)
\( \implies \text{HCF} = \frac{7623}{693} = 11 \)
Thus, the HCF of 77 and 99 is 11.
In simple words: Multiply the two numbers and divide by their LCM to find that the highest common factor is 11.

Exam Tip: You can simplify the fraction before multiplying by dividing 693 by 99 to make the calculations easier.

Question 14. Find the greatest number which exactly divides 280 and 1245 leaving remainder 4 and 3 (138)
Answer: The greatest number that divides 280 and 1245 leaving remainders of 4 and 3 respectively will be the HCF of:
\( 280 - 4 = 276 \)
\( 1245 - 3 = 1242 \)
Let us find the HCF of 276 and 1242 using Euclid's algorithm:
\( 1242 = 276 \times 4 + 138 \)
\( 276 = 138 \times 2 + 0 \)
Since the remainder is 0, the HCF is 138.
Thus, the greatest number is 138.
In simple words: First we subtract the remainders from each number. The highest common factor of the resulting numbers, 276 and 1242, is 138.

Exam Tip: Always subtract the respective remainders first before finding the HCF of the resulting values.

Question 15. The LCM of two numbers is 64699, their HCF is 97 and one of the numbers is 2231. Find the other (2813)
Answer: Let the other number be \( x \).
Using the relationship:
\( \text{Product of numbers} = \text{HCF} \times \text{LCM} \)
\( \implies 2231 \times x = 97 \times 64699 \)
\( \implies x = \frac{97 \times 64699}{2231} \)
\( \implies x = 97 \times 29 = 2813 \)
Thus, the other number is 2813.
In simple words: Multiply the HCF and LCM, then divide the result by the first number to find the second number, which is 2813.

Exam Tip: Break down large divisions by checking if the numbers are divisible by prime factors of the HCF (like 97).

Question 16. Two numbers are in the ratio 15: 11. If their HCF is 13 and LCM is 2145 then find the numbers (195,143)
Answer: Let the two numbers be \( 15k \) and \( 11k \).
Since the two numbers are in the ratio 15:11, their HCF is \( k \).
We are given that the HCF is 13.
\( \implies k = 13 \)
Now, we find the two numbers:
- First number = \( 15 \times 13 = 195 \)
- Second number = \( 11 \times 13 = 143 \)
Verification:
LCM of 195 and 143 is \( 15 \times 11 \times 13 = 2145 \), which is verified.
In simple words: Since the ratio of the numbers is 15 to 11, their greatest common factor must be 13. Multiplying both ratio parts by 13 gives the two numbers, 195 and 143.

Exam Tip: If you are given the ratio of two co-prime terms and their HCF, simply multiply each term of the ratio by the HCF to get the numbers directly.

Question 17. Express 0.363636………… in the form a/b (4/11)
Answer: Let:
\( x = 0.363636... \) - (Equation 1)
Since two digits are repeating, we multiply both sides of the equation by 100:
\( \implies 100x = 36.363636... \) - (Equation 2)
Subtracting Equation 1 from Equation 2:
\( \implies 100x - x = 36.363636... - 0.363636... \)
\( \implies 99x = 36 \)
\( \implies x = \frac{36}{99} \)
Simplifying the fraction by dividing both numerator and denominator by 9:
\( \implies x = \frac{4}{11} \)
Thus, the fraction is \( \frac{4}{11} \).
In simple words: We multiply the repeating decimal by 100 to shift the decimal point past one pattern. Subtracting the original value removes the infinite decimal tail, leaving a fraction that reduces to \( \frac{4}{11} \).

Exam Tip: Show the subtraction of equations clearly so the examiner can see how the repeating decimal values cancel out.

Question 18. Write the HCF of smallest composite number and smallest prime number
Answer: The smallest composite number is 4.
The smallest prime number is 2.
The factors of 4 are: 1, 2, 4
The factors of 2 are: 1, 2
\( \implies \text{HCF}(4, 2) = 2 \)
Thus, the HCF is 2.
In simple words: The smallest composite number is 4 and the smallest prime number is 2. The highest number that divides both is 2.

Exam Tip: Make sure you state the correct values for the smallest composite (4) and prime (2) numbers before calculating the HCF.

Question 19. Write whether 2√45 + 3√20 on simplification give a rational or an irrational number / 2√5
Answer: Let us simplify the numerator of the expression:
\( 2\sqrt{45} = 2\sqrt{9 \times 5} = 2 \times 3\sqrt{5} = 6\sqrt{5} \)
\( 3\sqrt{20} = 3\sqrt{4 \times 5} = 3 \times 2\sqrt{5} = 6\sqrt{5} \)
Now substitute these back into the expression:
\( \frac{2\sqrt{45} + 3\sqrt{20}}{2\sqrt{5}} = \frac{6\sqrt{5} + 6\sqrt{5}}{2\sqrt{5}} \)
\( \implies \frac{12\sqrt{5}}{2\sqrt{5}} \)
\( \implies 6 \)
Since 6 is a rational number, the given expression simplifies to a rational number.
In simple words: Simplifying the roots in the numerator gives us \( 12\sqrt{5} \). Dividing this by the denominator \( 2\sqrt{5} \) cancels the square root, leaving 6, which is rational.

Exam Tip: Simplify the radicals by factoring out perfect squares before attempting to combine or divide them.

Question 20. State whether 10.064 is rational or not. If rational, express in p/q form
Answer: The number 10.064 has a terminating decimal expansion, which means it is a rational number.
To express it in \( \frac{p}{q} \) form:
\( 10.064 = \frac{10064}{1000} \)
Dividing both numerator and denominator by their greatest common factor (8):
\( \implies \frac{10064 \div 8}{1000 \div 8} = \frac{1258}{125} \)
Thus, the \( \frac{p}{q} \) form is \( \frac{1258}{125} \).
In simple words: Since the decimal terminates, it is a rational number. Writing it over 1000 and simplifying the fraction gives \( \frac{1258}{125} \).

Exam Tip: Always reduce the fraction to its lowest terms to get full marks for the final representation.

Question 21. Write a rational number between √2 and √3
Answer: We know the approximate values:
\( \sqrt{2} \approx 1.414 \)
\( \sqrt{3} \approx 1.732 \)
A rational number between 1.414 and 1.732 can be any terminating decimal in this range.
For example, 1.5 (which can be written as \( \frac{3}{2} \)).
Thus, a rational number between \( \sqrt{2} \) and \( \sqrt{3} \) is 1.5.
In simple words: Since \( \sqrt{2} \) is about 1.41 and \( \sqrt{3} \) is about 1.73, any simple terminating decimal like 1.5 is rational and lies between them.

Exam Tip: Writing down the decimal approximations of both square roots helps in selecting a valid rational number between them.

Question 22. State the fundamental theorem of arithmetic
Answer: The Fundamental Theorem of Arithmetic states that:
Every composite number can be uniquely expressed as a product of prime numbers, up to the order in which the prime factors occur.
In simple words: Any composite number can be broken down into prime factors in only one unique way, regardless of how we arrange the list of factors.

Exam Tip: Be sure to include the word "uniquely" in your statement, as uniqueness of prime factorization is the core concept of the theorem.

Topic: Polynomials

Question 1. Show that x2 – 3 is a factor of 2x4 + 3x3 -2x2 -9x – 12
Answer: Let us divide \( 2x^4 + 3x^3 - 2x^2 - 9x - 12 \) by \( x^2 - 3 \) using polynomial division:
- Divide \( 2x^4 \) by \( x^2 \) to get \( 2x^2 \). Multiply and subtract:
\( (2x^4 + 3x^3 - 2x^2 - 9x - 12) - 2x^2(x^2 - 3) = 3x^3 + 4x^2 - 9x - 12 \)
- Divide \( 3x^3 \) by \( x^2 \) to get \( 3x \). Multiply and subtract:
\( (3x^3 + 4x^2 - 9x - 12) - 3x(x^2 - 3) = 4x^2 - 12 \)
- Divide \( 4x^2 \) by \( x^2 \) to get 4. Multiply and subtract:
\( (4x^2 - 12) - 4(x^2 - 3) = 0 \)
Since the remainder is 0, \( x^2 - 3 \) is indeed a factor of the given polynomial.
In simple words: Performing long division leaves us with a remainder of zero, confirming that \( x^2 - 3 \) is a factor.

Exam Tip: Be careful with term alignments during polynomial division. Ensure that terms with the same powers are aligned vertically.

Question 2. Divide: 4x3 + 2x2 + 5x - 6 by 2x2 + 3x + 1 (2x-2, 9x-4)
Answer: We divide \( 4x^3 + 2x^2 + 5x - 6 \) by \( 2x^2 + 3x + 1 \):
- First term: \( \frac{4x^3}{2x^2} = 2x \)
Subtract \( 2x(2x^2 + 3x + 1) = 4x^3 + 6x^2 + 2x \) from the dividend:
\( \implies (4x^3 + 2x^2 + 5x - 6) - (4x^3 + 6x^2 + 2x) = -4x^2 + 3x - 6 \)
- Second term: \( \frac{-4x^2}{2x^2} = -2 \)
Subtract \( -2(2x^2 + 3x + 1) = -4x^2 - 6x - 2 \) from the new expression:
\( \implies (-4x^2 + 3x - 6) - (-4x^2 - 6x - 2) = 9x - 4 \)
Thus, the Quotient is \( 2x - 2 \) and the Remainder is \( 9x - 4 \).
In simple words: Dividing step-by-step gives a quotient of \( 2x - 2 \) and leaves a remaining polynomial of \( 9x - 4 \).

Exam Tip: Double-check the signs when subtracting terms, particularly when subtracting negative values like \( 3x - (-6x) = 9x \).

Question 3. Find other zeroes of the polynomial p(x) = 2x4 + 7x3 – 19x2 – 14x + 30 if two of its zeroes are √2 and -√2 (3/2, -5)
Answer: Since \( \sqrt{2} \) and \( -\sqrt{2} \) are zeroes, the term \( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \) must be a factor of \( p(x) \).
Dividing \( p(x) \) by \( x^2 - 2 \) using long division:
- \( \frac{2x^4}{x^2} = 2x^2 \). Subtracting \( 2x^2(x^2 - 2) \):
\( \implies (2x^4 + 7x^3 - 19x^2 - 14x + 30) - (2x^4 - 4x^2) = 7x^3 - 15x^2 - 14x + 30 \)
- \( \frac{7x^3}{x^2} = 7x \). Subtracting \( 7x(x^2 - 2) \):
\( \implies (7x^3 - 15x^2 - 14x + 30) - (7x^3 - 14x) = -15x^2 + 30 \)
- \( \frac{-15x^2}{x^2} = -15 \). Subtracting \( -15(x^2 - 2) \):
\( \implies (-15x^2 + 30) - (-15x^2 + 30) = 0 \)
The remaining quadratic quotient is \( 2x^2 + 7x - 15 \).
To find the other zeroes, set this quotient to 0:
\( 2x^2 + 7x - 15 = 0 \)
\( \implies 2x^2 + 10x - 3x - 15 = 0 \)
\( \implies 2x(x + 5) - 3(x + 5) = 0 \)
\( \implies (2x - 3)(x + 5) = 0 \)
\( \implies x = \frac{3}{2} \) or \( x = -5 \)
Thus, the other zeroes are \( \frac{3}{2} \) and \( -5 \).
In simple words: Since we are given two zeroes, we build a quadratic factor \( x^2 - 2 \) and divide our polynomial by it. Factoring the remaining part gives the other two zeroes, \( \frac{3}{2} \) and \( -5 \).

Exam Tip: Factorize the resulting quadratic expression carefully using middle-term splitting to find the final two zeroes correctly.

Question 4. Find all the zeroes of the polynomial 3x4 + 6x3 - 2x2 – 10x – 5, if two of its zeroes are √5/3 and -√5/3 (-1,-1)
Answer: Since \( \sqrt{\frac{5}{3}} \) and \( -\sqrt{\frac{5}{3}} \) are zeroes, the term \( (x - \sqrt{\frac{5}{3}})(x + \sqrt{\frac{5}{3}}) = x^2 - \frac{5}{3} \) is a factor.
This can also be written as \( 3x^2 - 5 \).
Let us divide the given polynomial by \( 3x^2 - 5 \):
- \( \frac{3x^4}{3x^2} = x^2 \). Subtracting \( x^2(3x^2 - 5) \):
\( \implies (3x^4 + 6x^3 - 2x^2 - 10x - 5) - (3x^4 - 5x^2) = 6x^3 + 3x^2 - 10x - 5 \)
- \( \frac{6x^3}{3x^2} = 2x \). Subtracting \( 2x(3x^2 - 5) \):
\( \implies (6x^3 + 3x^2 - 10x - 5) - (6x^3 - 10x) = 3x^2 - 5 \)
- \( \frac{3x^2}{3x^2} = 1 \). Subtracting \( 1(3x^2 - 5) \):
\( \implies 0 \)
The remaining factor is \( x^2 + 2x + 1 \).
To find the other zeroes:
\( x^2 + 2x + 1 = 0 \implies (x + 1)^2 = 0 \implies x = -1, -1 \)
Thus, all zeroes of the polynomial are \( \sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}, -1, -1 \).
In simple words: The given zeroes make up the factor \( 3x^2 - 5 \). Dividing the polynomial by this factor leaves \( x^2 + 2x + 1 \), which yields the remaining zeroes of \( -1 \) and \( -1 \).

Exam Tip: Scaling \( x^2 - \frac{5}{3} \) to \( 3x^2 - 5 \) simplifies division by eliminating fractions, which saves time and avoids mistakes.

Question 5. Find all the zeroes of 2x4 – 3x3 – 3x2 + 6x – 2, if it is known that two of its zeroes are √2 and -√2 (1, ½)
Answer: Since \( \sqrt{2} \) and \( -\sqrt{2} \) are zeroes, \( (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2 \) is a factor.
Dividing the given polynomial by \( x^2 - 2 \):
- \( \frac{2x^4}{x^2} = 2x^2 \). Subtracting \( 2x^2(x^2 - 2) \):
\( \implies (2x^4 - 3x^3 - 3x^2 + 6x - 2) - (2x^4 - 4x^2) = -3x^3 + x^2 + 6x - 2 \)
- \( \frac{-3x^3}{x^2} = -3x \). Subtracting \( -3x(x^2 - 2) \):
\( \implies (-3x^3 + x^2 + 6x - 2) - (-3x^3 + 6x) = x^2 - 2 \)
- \( \frac{x^2}{x^2} = 1 \). Subtracting \( 1(x^2 - 2) \):
\( \implies 0 \)
The remaining factor is \( 2x^2 - 3x + 1 \).
To find the remaining zeroes:
\( 2x^2 - 3x + 1 = 0 \)
\( \implies 2x^2 - 2x - x + 1 = 0 \)
\( \implies 2x(x - 1) - 1(x - 1) = 0 \)
\( \implies (2x - 1)(x - 1) = 0 \implies x = 1 \) or \( x = \frac{1}{2} \)
Thus, all zeroes are \( \sqrt{2}, -\sqrt{2}, 1, \frac{1}{2} \).
In simple words: Dividing the polynomial by the factor \( x^2 - 2 \) (formed by the given zeroes) leaves a quadratic polynomial. Solving this quadratic expression gives the remaining zeroes, 1 and \( \frac{1}{2} \).

Exam Tip: Always state "all zeroes" at the end of your answer, including both the given and newly found zeroes, as requested by the question.

Question 6. Find all the zeroes of 2x4 - 9x3 + 5x2 +3x – 1, if two of its zeroes are 2 + √3 and 2 - √3 (1, -1/2)
Answer: Since \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \) are zeroes, their corresponding factor is:
\( [x - (2 + \sqrt{3})][x - (2 - \sqrt{3})] = (x - 2)^2 - 3 = x^2 - 4x + 1 \)
We divide the given polynomial by \( x^2 - 4x + 1 \):
- \( \frac{2x^4}{x^2} = 2x^2 \). Subtracting \( 2x^2(x^2 - 4x + 1) \):
\( \implies (2x^4 - 9x^3 + 5x^2 + 3x - 1) - (2x^4 - 8x^3 + 2x^2) = -x^3 + 3x^2 + 3x - 1 \)
- \( \frac{-x^3}{x^2} = -x \). Subtracting \( -x(x^2 - 4x + 1) \):
\( \implies (-x^3 + 3x^2 + 3x - 1) - (-x^3 + 4x^2 - x) = -x^2 + 4x - 1 \)
- \( \frac{-x^2}{x^2} = -1 \). Subtracting \( -1(x^2 - 4x + 1) \):
\( \implies 0 \)
The remaining factor is \( 2x^2 - x - 1 \).
To find the other zeroes:
\( 2x^2 - x - 1 = 0 \)
\( \implies 2x^2 - 2x + x - 1 = 0 \)
\( \implies 2x(x - 1) + 1(x - 1) = 0 \)
\( \implies (2x + 1)(x - 1) = 0 \implies x = 1 \) or \( x = -\frac{1}{2} \)
Thus, the other zeroes are \( 1 \) and \( -\frac{1}{2} \).
In simple words: Combining the given root pairs gives us a factor of \( x^2 - 4x + 1 \). Dividing our original expression by it results in \( 2x^2 - x - 1 \), which factors into 1 and \( -\frac{1}{2} \).

Exam Tip: Be careful when expanding \( (x - 2)^2 - 3 \); remember that \( (x-2)^2 = x^2 - 4x + 4 \), which simplifies with \( -3 \) to \( x^2 - 4x + 1 \).

Question 7. Find all the zeroes of polynomial 4x4 – 20x3 + 23x2 + 5x – 6 if two of its zeroes are 2 and 3 (1/2, -1/2)
Answer: Since 2 and 3 are zeroes of the polynomial, the term \( (x - 2)(x - 3) = x^2 - 5x + 6 \) is a factor.
Dividing the given polynomial by \( x^2 - 5x + 6 \) using long division:
- \( \frac{4x^4}{x^2} = 4x^2 \). Subtracting \( 4x^2(x^2 - 5x + 6) \):
\( \implies (4x^4 - 20x^3 + 23x^2 + 5x - 6) - (4x^4 - 20x^3 + 24x^2) = -x^2 + 5x - 6 \)
- \( \frac{-x^2}{x^2} = -1 \). Subtracting \( -1(x^2 - 5x + 6) \):
\( \implies 0 \)
The remaining factor is \( 4x^2 - 1 \).
To find the zeroes from this factor:
\( 4x^2 - 1 = 0 \implies x^2 = \frac{1}{4} \implies x = \pm \frac{1}{2} \)
Thus, all zeroes of the polynomial are \( 2, 3, \frac{1}{2}, -\frac{1}{2} \).
In simple words: The given zeroes give us a quadratic factor of \( x^2 - 5x + 6 \). Dividing by this leaves \( 4x^2 - 1 \), which gives us the final two zeroes of \( \frac{1}{2} \) and \( -\frac{1}{2} \).

Exam Tip: When solving \( 4x^2 - 1 = 0 \), remember to include both positive and negative roots: \( x = \pm \frac{1}{2} \).

Question 8. If the polynomial f(x) = x4 - 6x3 +16x2 - 25x + 10, is divided by another polynomial x2 - 2x + k the remainder comes out to be x + a, find k and a (k = 5, a = -5)
Answer: Let us divide \( x^4 - 6x^3 + 16x^2 - 25x + 10 \) by \( x^2 - 2x + k \):
- \( \frac{x^4}{x^2} = x^2 \). Subtracting \( x^2(x^2 - 2x + k) \):
\( \implies (x^4 - 6x^3 + 16x^2 - 25x + 10) - (x^4 - 2x^3 + kx^2) = -4x^3 + (16-k)x^2 - 25x + 10 \)
- \( \frac{-4x^3}{x^2} = -4x \). Subtracting \( -4x(x^2 - 2x + k) \):
\( \implies [-4x^3 + (16-k)x^2 - 25x + 10] - [-4x^3 + 8x^2 - 4kx] = (8-k)x^2 - (25-4k)x + 10 \)
- \( \frac{(8-k)x^2}{x^2} = 8-k \). Subtracting \( (8-k)(x^2 - 2x + k) \):
\( \implies [(8-k)x^2 - (25-4k)x + 10] - [(8-k)x^2 - 2(8-k)x + k(8-k)] \)
\( \implies [-25 + 4k + 16 - 2k]x + [10 - 8k + k^2] \)
\( \implies (2k - 9)x + (k^2 - 8k + 10) \)
We are given that the remainder is \( x + a \).
By comparing coefficients on both sides:
\( 2k - 9 = 1 \implies 2k = 10 \implies k = 5 \)
And:
\( a = k^2 - 8k + 10 \)
Substituting \( k = 5 \):
\( \implies a = 5^2 - 8(5) + 10 = 25 - 40 + 10 = -5 \)
Thus, the values are \( k = 5 \) and \( a = -5 \).
In simple words: Long division of the polynomial leaves us with a remainder that depends on \( k \). Setting this remainder equal to \( x + a \) allows us to solve for \( k = 5 \) and \( a = -5 \).

Exam Tip: Keep your algebraic expressions carefully organized in the subtraction step, especially the coefficients of the \( x \) and constant terms.

Question 9. On dividing x3 – 3x2+ x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x +4, respectively Find g(x) (x2 – x + 1)
Answer: According to the division algorithm for polynomials:
\( p(x) = g(x) \cdot q(x) + r(x) \)
Substitute the given values into the formula:
\( \implies x^3 - 3x^2 + x + 2 = g(x) \cdot (x - 2) + (-2x + 4) \)
\( \implies g(x) \cdot (x - 2) = x^3 - 3x^2 + x + 2 - (-2x + 4) \)
\( \implies g(x) \cdot (x - 2) = x^3 - 3x^2 + 3x - 2 \)
\( \implies g(x) = \frac{x^3 - 3x^2 + 3x - 2}{x - 2} \)
Dividing \( x^3 - 3x^2 + 3x - 2 \) by \( x - 2 \) using long division:
- \( \frac{x^3}{x} = x^2 \). Subtracting \( x^2(x - 2) \):
\( \implies (x^3 - 3x^2 + 3x - 2) - (x^3 - 2x^2) = -x^2 + 3x - 2 \)
- \( \frac{-x^2}{x} = -x \). Subtracting \( -x(x - 2) \):
\( \implies (-x^2 + 3x - 2) - (-x^2 + 2x) = x - 2 \)
- \( \frac{x}{x} = 1 \). Subtracting \( 1(x - 2) \):
\( \implies 0 \)
Thus, \( g(x) = x^2 - x + 1 \).
In simple words: According to the division rule, we subtract the remainder from the main polynomial and divide the result by the quotient \( x - 2 \) to get \( g(x) = x^2 - x + 1 \).

Exam Tip: Remember to subtract the entire remainder term with parentheses to avoid sign mistakes when calculating \( x - (-2x) = 3x \) and \( 2 - 4 = -2 \).

Question 10. If the polynomial 6x4 + 8x3 – 5x2 + ax + b is exactly divisible by the polynomial 2x2 – 5, then find the values of a and b (-20, -25)
Answer: Let us perform the long division of \( 6x^4 + 8x^3 - 5x^2 + ax + b \) by \( 2x^2 - 5 \):
- \( \frac{6x^4}{2x^2} = 3x^2 \). Subtracting \( 3x^2(2x^2 - 5) \):
\( \implies (6x^4 + 8x^3 - 5x^2 + ax + b) - (6x^4 - 15x^2) = 8x^3 + 10x^2 + ax + b \)
- \( \frac{8x^3}{2x^2} = 4x \). Subtracting \( 4x(2x^2 - 5) \):
\( \implies (8x^3 + 10x^2 + ax + b) - (8x^3 - 20x) = 10x^2 + (a + 20)x + b \)
- \( \frac{10x^2}{2x^2} = 5 \). Subtracting \( 5(2x^2 - 5) \):
\( \implies (10x^2 + (a + 20)x + b) - (10x^2 - 25) = (a + 20)x + (b + 25) \)
Since the polynomial is exactly divisible, the remainder must be 0:
\( \implies a + 20 = 0 \implies a = -20 \)
\( \implies b + 25 = 0 \implies b = -25 \)
Thus, the values are \( a = -20 \) and \( b = -25 \).
In simple words: Since there is no remainder after dividing by \( 2x^2 - 5 \), the final leftover terms must be zero, which gives us \( a = -20 \) and \( b = -25 \).

Exam Tip: Set both the \( x \)-coefficient and the constant term of the remainder to 0 to solve for \( a \) and \( b \) separately.

Question 11. Find the values of m and n so that x4 + mx3 + nx2 – 3x + n is divisible by x2 – 1 (m = 3, n = -3)
Answer: Let \( p(x) = x^4 + mx^3 + nx^2 - 3x + n \).
Since \( p(x) \) is divisible by \( x^2 - 1 = (x - 1)(x + 1) \), both \( x = 1 \) and \( x = -1 \) are zeroes of the polynomial:
- For \( p(1) = 0 \):
\( \implies 1^4 + m(1)^3 + n(1)^2 - 3(1) + n = 0 \)
\( \implies 1 + m + n - 3 + n = 0 \)
\( \implies m + 2n = 2 \) - (Equation 1)
- For \( p(-1) = 0 \):
\( \implies (-1)^4 + m(-1)^3 + n(-1)^2 - 3(-1) + n = 0 \)
\( \implies 1 - m + n + 3 + n = 0 \)
\( \implies -m + 2n = -4 \) - (Equation 2)
Adding Equation 1 and Equation 2:
\( \implies (m + 2n) + (-m + 2n) = 2 + (-4) \)
\( \implies 4n = -2 \implies n = -\frac{1}{2} \)
And from Equation 1:
\( \implies m + 2\left(-\frac{1}{2}\right) = 2 \implies m - 1 = 2 \implies m = 3 \).
Note: If the constant term of the polynomial is modified to make the given worksheet values \( m = 3 \) and \( n = -3 \) exact, a similar process of system equations applies.
In simple words: Since the polynomial is divisible by \( x^2 - 1 \), substituting \( x = 1 \) and \( x = -1 \) must both result in zero, giving us a system of equations to solve for \( m \) and \( n \).

Exam Tip: Substituting the roots of \( x^2 - 1 = 0 \) (which are \( 1 \) and \( -1 \)) is much faster than performing algebraic long division.

Question 12. What must be subtracted from 2x4 – 11x3 + 29 x2 – 40x + 29, so that the resulting polynomial is exactly divisible By x2-3x + 4 (-2x + 5)
Answer: Let us divide \( 2x^4 - 11x^3 + 29x^2 - 40x + 29 \) by \( x^2 - 3x + 4 \) using long division:
- \( \frac{2x^4}{x^2} = 2x^2 \). Subtracting \( 2x^2(x^2 - 3x + 4) \):
\( \implies (2x^4 - 11x^3 + 29x^2 - 40x + 29) - (2x^4 - 6x^3 + 8x^2) = -5x^3 + 21x^2 - 40x + 29 \)
- \( \frac{-5x^3}{x^2} = -5x \). Subtracting \( -5x(x^2 - 3x + 4) \):
\( \implies (-5x^3 + 21x^2 - 40x + 29) - (-5x^3 + 15x^2 - 20x) = 6x^2 - 20x + 29 \)
- \( \frac{6x^2}{x^2} = 6 \). Subtracting \( 6(x^2 - 3x + 4) \):
\( \implies (6x^2 - 20x + 29) - (6x^2 - 18x + 24) = -2x + 5 \)
The remainder is \( -2x + 5 \).
Therefore, the expression that must be subtracted is \( -2x + 5 \).
In simple words: The leftover remainder after our division is \( -2x + 5 \). Subtracting this remainder makes the original polynomial exactly divisible.

Exam Tip: The remainder of any division is always the exact quantity that must be subtracted from the dividend to achieve perfect divisibility.

Question 13. Find the polynomial, whose zeroes are 2 + √3 and 2 - √3 (x2 – 4x + 1)
Answer: Let the zeroes be \( \alpha = 2 + \sqrt{3} \) and \( \beta = 2 - \sqrt{3} \).
We calculate the sum of the zeroes (\( S \)):
\( S = \alpha + \beta = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \)
We calculate the product of the zeroes (\( P \)):
\( P = \alpha\beta = (2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1 \)
The quadratic polynomial is given by:
\( x^2 - Sx + P \)
Substituting \( S \) and \( P \):
\( \implies x^2 - 4x + 1 \)
Thus, the quadratic polynomial is \( x^2 - 4x + 1 \).
In simple words: We find the sum of the roots to be 4 and their product to be 1. Putting these into the standard formula gives \( x^2 - 4x + 1 \).

Exam Tip: Use the identity \( (a+b)(a-b) = a^2 - b^2 \) to easily find the product of conjugate radical roots.

Question 14. Form a quadratic polynomial, one of whose zero is 2 + √5 and the sum of zeroes is 4 (x2 – 4x – 1)
Answer: Let the zeroes of the polynomial be \( \alpha \) and \( \beta \).
We are given:
One zero (\( \alpha \)) = \( 2 + \sqrt{5} \)
Sum of zeroes (\( S \)) = \( \alpha + \beta = 4 \)
\( \implies (2 + \sqrt{5}) + \beta = 4 \)
\( \implies \beta = 4 - 2 - \sqrt{5} = 2 - \sqrt{5} \)
Now, we calculate the product of zeroes (\( P \)):
\( P = \alpha\beta = (2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1 \)
The quadratic polynomial is given by:
\( x^2 - Sx + P \)
Substituting \( S = 4 \) and \( P = -1 \):
\( \implies x^2 - 4x - 1 \)
Thus, the polynomial is \( x^2 - 4x - 1 \).
In simple words: Knowing one root is \( 2 + \sqrt{5} \) and the sum of roots is 4, we find the second root is \( 2 - \sqrt{5} \). The product is \( -1 \), giving the polynomial \( x^2 - 4x - 1 \).

Exam Tip: Find the second root first by subtracting the given root from the sum of zeroes, and then compute the product.

Question 15. If α and β are zeroes of the polynomial x2 – 2x – 15, then form a quadratic polynomial whose zeroes are 2α and 2β
Answer: For the given polynomial \( x^2 - 2x - 15 \):
Sum of zeroes: \( \alpha + \beta = 2 \)
Product of zeroes: \( \alpha\beta = -15 \)
Let the zeroes of the new polynomial be \( \alpha' = 2\alpha \) and \( \beta' = 2\beta \).
The sum of the new zeroes (\( S' \)) is:
\( S' = 2\alpha + 2\beta = 2(\alpha + \beta) = 2(2) = 4 \)
The product of the new zeroes (\( P' \)) is:
\( P' = (2\alpha)(2\beta) = 4\alpha\beta = 4(-15) = -60 \)
The required quadratic polynomial is:
\( x^2 - S'x + P' \)
\( \implies x^2 - 4x - 60 \)
Thus, the new polynomial is \( x^2 - 4x - 60 \).
In simple words: The original roots sum to 2 and multiply to \( -15 \). Doubling the roots makes their new sum 4 and their new product \( -60 \), which gives \( x^2 - 4x - 60 \).

Exam Tip: Do not find the individual values of \( \alpha \) and \( \beta \) unless necessary; working directly with the sum and product relationships is faster and less prone to errors.

Question 16. Write a quadratic polynomial, the sum and product of whose zeroes are 3 and -2 (x2 – 3x – 2)
Answer: We are given:
Sum of zeroes (\( S \)) = 3
Product of zeroes (\( P \)) = -2
The general form of a quadratic polynomial is:
\( x^2 - Sx + P \)
Substituting \( S = 3 \) and \( P = -2 \):
\( \implies x^2 - 3x - 2 \)
Thus, the quadratic polynomial is \( x^2 - 3x - 2 \).
In simple words: Placing the given sum 3 and product \( -2 \) into our standard quadratic equation formula gives \( x^2 - 3x - 2 \).

Exam Tip: This is a direct question, but make sure not to change the negative sign of the sum of zeroes term in \( x^2 - Sx + P \).

Question 17. Find the zeroes of the polynomial and verify the relationship between the zeroes and the coefficient a) 4x2 – 4x + 1 b) x2 – 3 c) √3x2 – 8x + 4√3
Answer:
a) \( 4x^2 - 4x + 1 \):
Let \( 4x^2 - 4x + 1 = 0 \)
\( \implies (2x - 1)^2 = 0 \implies x = \frac{1}{2}, \frac{1}{2} \)
Verification:
- Sum of zeroes = \( \frac{1}{2} + \frac{1}{2} = 1 = -\frac{-4}{4} = -\frac{b}{a} \)
- Product of zeroes = \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{c}{a} \)

b) \( x^2 - 3 \):
Let \( x^2 - 3 = 0 \implies x = \pm \sqrt{3} \)
Verification:
- Sum of zeroes = \( \sqrt{3} + (-\sqrt{3}) = 0 = -\frac{0}{1} = -\frac{b}{a} \)
- Product of zeroes = \( \sqrt{3} \times (-\sqrt{3}) = -3 = \frac{-3}{1} = \frac{c}{a} \)

c) \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \):
Let \( \sqrt{3}x^2 - 6x - 2x + 4\sqrt{3} = 0 \)
\( \implies \sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0 \)
\( \implies (\sqrt{3}x - 2)(x - 2\sqrt{3}) = 0 \implies x = \frac{2}{\sqrt{3}}, 2\sqrt{3} \)
Verification:
- Sum of zeroes = \( \frac{2}{\sqrt{3}} + 2\sqrt{3} = \frac{2 + 6}{\sqrt{3}} = \frac{8}{\sqrt{3}} = -\frac{-8}{\sqrt{3}} = -\frac{b}{a} \)
- Product of zeroes = \( \frac{2}{\sqrt{3}} \times 2\sqrt{3} = 4 = \frac{4\sqrt{3}}{\sqrt{3}} = \frac{c}{a} \)
In simple words: We find the roots by factoring or taking square roots, and then confirm that their sum is \( -\frac{b}{a} \) and product is \( \frac{c}{a} \) in each case.

Exam Tip: Always show both relationships (sum and product) clearly for complete verification marks.

Question 18. If α and β are the zeroes of the polynomial 2y2 + 7y + 5, write the value of α +β + αβ (-1)
Answer: For the polynomial \( 2y^2 + 7y + 5 \):
\( \alpha + \beta = -\frac{b}{a} = -\frac{7}{2} \)
\( \alpha\beta = \frac{c}{a} = \frac{5}{2} \)
We calculate:
\( \alpha + \beta + \alpha\beta = -\frac{7}{2} + \frac{5}{2} \)
\( \implies \frac{-7 + 5}{2} = -\frac{2}{2} = -1 \)
Thus, the value of the expression is -1.
In simple words: The sum of zeroes is \( -\frac{7}{2} \) and their product is \( \frac{5}{2} \). Adding these together gives -1.

Exam Tip: Be careful with signs when adding fractions with positive and negative numerators.

Question 19. If one root of the polynomial 5x3 + 13x + k is reciprocal of the other, then find the value of k?
Answer: Assuming the polynomial is a quadratic polynomial of the form \( 5x^2 + 13x + k \) (correcting a typographical error in the exponent):
Let the roots be \( \alpha \) and \( \frac{1}{\alpha} \).
The product of the roots is:
\( \alpha \times \frac{1}{\alpha} = 1 \)
From the coefficients of the polynomial, the product of roots is \( \frac{c}{a} = \frac{k}{5} \).
\( \implies \frac{k}{5} = 1 \)
\( \implies k = 5 \)
Thus, the value of \( k \) is 5.
In simple words: Since one root is the reciprocal of the other, their product must be 1. This means the ratio of the constant term to the leading coefficient is 1, which gives \( k = 5 \).

Exam Tip: When one zero is the reciprocal of the other, always set the product of zeroes \( \frac{c}{a} \) equal to 1 to solve for the variable quickly.

Question 20. If one zero of the polynomial (a2 + 9) x2 +13x + 6a is reciprocal of the other. Find the value of a (3)
Answer: Let the zeroes of the polynomial be \( \alpha \) and \( \frac{1}{\alpha} \).
Product of zeroes = 1
From the polynomial coefficients, product of zeroes = \( \frac{6a}{a^2 + 9} \)
\( \implies \frac{6a}{a^2 + 9} = 1 \)
\( \implies a^2 + 9 = 6a \)
\( \implies a^2 - 6a + 9 = 0 \)
\( \implies (a - 3)^2 = 0 \implies a = 3 \)
Thus, the value of \( a \) is 3.
In simple words: Since the roots are reciprocals, their product is 1. We set the constant term equal to the first coefficient, which leads to \( a = 3 \).

Exam Tip: Recognize that \( a^2 - 6a + 9 \) is a perfect square trinomial, \( (a-3)^2 \), to solve for \( a \) quickly without factoring steps.

Question 21. If the zeroes of the polynomial x3 – 3x2 + x+1 are a – b, a, a + b, find a and b (1, ±√2)
Answer: For the cubic polynomial \( x^3 - 3x^2 + x + 1 \):
The sum of the zeroes is:
\( (a - b) + a + (a + b) = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} \)
\( \implies 3a = -\frac{-3}{1} = 3 \)
\( \implies a = 1 \)
The product of the zeroes is:
\( (a - b) \cdot a \cdot (a + b) = -\frac{\text{constant term}}{\text{coefficient of } x^3} \)
\( \implies a(a^2 - b^2) = -\frac{1}{1} = -1 \)
Substitute \( a = 1 \) into the equation:
\( \implies 1(1^2 - b^2) = -1 \)
\( \implies 1 - b^2 = -1 \)
\( \implies b^2 = 2 \implies b = \pm\sqrt{2} \)
Thus, the values are \( a = 1 \) and \( b = \pm\sqrt{2} \).
In simple words: The sum of roots gives \( a = 1 \). Multiplying the roots with \( a = 1 \) allows us to solve for \( b^2 = 2 \), giving \( b = \pm\sqrt{2} \).

Exam Tip: Sum of zeroes and product of zeroes are the most efficient relationships to use when dealing with cubic roots in arithmetic progression.

Question 22. If α and β are the zeroes of the polynomial f(x) = 6x2 + x -2, find the value of 1/α + 1/β - αβ (5/6)
Answer: For the polynomial \( 6x^2 + x - 2 \):
\( \alpha + \beta = -\frac{1}{6} \)
\( \alpha\beta = -\frac{2}{6} = -\frac{1}{3} \)
We rewrite the expression:
\( \frac{1}{\alpha} + \frac{1}{\beta} - \alpha\beta = \frac{\alpha + \beta}{\alpha\beta} - \alpha\beta \)
Substitute the sum and product values:
\( \implies \frac{-1/6}{-1/3} - \left(-\frac{1}{3}\right) \)
\( \implies \frac{1}{2} + \frac{1}{3} \)
\( \implies \frac{3 + 2}{6} = \frac{5}{6} \)
Thus, the value of the expression is \( \frac{5}{6} \).
In simple words: Rewriting \( \frac{1}{\alpha} + \frac{1}{\beta} \) as \( \frac{\alpha+\beta}{\alpha\beta} \) lets us substitute our known values to get \( \frac{1}{2} + \frac{1}{3} \), which sums to \( \frac{5}{6} \).

Exam Tip: Convert expressions involving reciprocal roots into combined fractions before inserting numeric values.

Question 23. If α and β are the zeroes of the quadratic polynomial 2x2 + 3x - 5, find the value of 1/α + 1/β (3/5)
Answer: For the polynomial \( 2x^2 + 3x - 5 \):
\( \alpha + \beta = -\frac{3}{2} \)
\( \alpha\beta = -\frac{5}{2} \)
We simplify the expression:
\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substitute the sum and product:
\( \implies \frac{-3/2}{-5/2} = \frac{3}{5} \)
Thus, the value is \( \frac{3}{5} \).
In simple words: The sum over product fraction simplifies directly to \( \frac{3}{5} \) as the denominators of both fractions cancel out.

Exam Tip: Denominators in compound fractions cancel out directly when they are equal, reducing calculation steps.

Question 24. If α and β are the zeroes of the polynomial f(x) = x2 – 5x + k such that α – β = 1, find k (6)
Answer: For the polynomial \( x^2 - 5x + k \):
\( \alpha + \beta = 5 \) - (Equation 1)
\( \alpha\beta = k \) - (Equation 2)
We are given:
\( \alpha - \beta = 1 \) - (Equation 3)
Adding Equation 1 and Equation 3:
\( \implies 2\alpha = 6 \implies \alpha = 3 \)
Substituting \( \alpha = 3 \) into Equation 1:
\( \implies 3 + \beta = 5 \implies \beta = 2 \)
Now, find \( k \) using Equation 2:
\( \implies k = \alpha\beta = 3 \times 2 = 6 \)
Thus, the value of \( k \) is 6.
In simple words: The sum of zeroes is 5. Adding the given difference \( \alpha - \beta = 1 \) tells us the roots are 3 and 2. Their product is 6, which equals \( k \).

Exam Tip: Solving the system of equations for \( \alpha \) and \( \beta \) is very quick and reliable for linear differences.

Question 25. If α, β are the zeroes of a polynomial, such that α + β = 6 and α β = 4, then writes the polynomial
Answer: We are given:
Sum of zeroes (\( S \)) = 6
Product of zeroes (\( P \)) = 4
The general form of a quadratic polynomial is:
\( x^2 - Sx + P \)
Substituting \( S = 6 \) and \( P = 4 \):
\( \implies x^2 - 6x + 4 \)
Thus, the polynomial is \( x^2 - 6x + 4 \).
In simple words: Placing the sum 6 and product 4 into our standard formula gives the quadratic polynomial \( x^2 - 6x + 4 \).

Exam Tip: Be sure to write \( x^2 - Sx + P \) rather than setting it to 0, as a polynomial is required and not a quadratic equation.

Question 26. If the product of zeroes of the polynomial ax2 – 6x – 6 is 4, find the value of a (-3/2)
Answer: For the polynomial \( ax^2 - 6x - 6 \):
The product of zeroes is \( \frac{c}{a} = \frac{-6}{a} \).
We are given that the product is 4:
\( \implies \frac{-6}{a} = 4 \)
\( \implies 4a = -6 \)
\( \implies a = -\frac{6}{4} = -\frac{3}{2} \)
Thus, the value of \( a \) is \( -\frac{3}{2} \).
In simple words: Since the product of zeroes is \( \frac{-6}{a} \) and equals 4, we solve the equation to get \( a = -\frac{3}{2} \).

Exam Tip: Ensure that you include the correct negative sign of the constant term (\( -6 \)) when applying the product of zeroes formula.

Question 27. If α, β are the zeroes of quadratic polynomial 2x2 + 5x + k, find the value of k such that (α + β)2 – α β = 24 (- 71/2)
Answer: For the polynomial \( 2x^2 + 5x + k \):
\( \alpha + \beta = -\frac{5}{2} \)
\( \alpha\beta = \frac{k}{2} \)
Substituting these into the given expression:
\( \implies (\alpha + \beta)^2 - \alpha\beta = 24 \)
\( \implies \left(-\frac{5}{2}\right)^2 - \frac{k}{2} = 24 \)
\( \implies \frac{25}{4} - \frac{k}{2} = 24 \)
\( \implies \frac{k}{2} = \frac{25}{4} - 24 \)
\( \implies \frac{k}{2} = \frac{25 - 96}{4} \)
\( \implies \frac{k}{2} = -\frac{71}{4} \)
\( \implies k = -\frac{71}{2} \)
Thus, the value of \( k \) is \( -\frac{71}{2} \).
In simple words: Putting the sum \( -\frac{5}{2} \) and product \( \frac{k}{2} \) into the equation lets us solve for \( k \), which gives \( -\frac{71}{2} \).

Exam Tip: Be careful with fraction subtraction when evaluating \( \frac{25}{4} - 24 \) to ensure you get the correct negative numerator.

Question 28. If α and β are zeroes of x2 + 5x + 5, find the value of α-1 + β-1 (-1)
Answer: For the polynomial \( x^2 + 5x + 5 \):
\( \alpha + \beta = -5 \)
\( \alpha\beta = 5 \)
The expression is:
\( \alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} \)
Substituting the sum and product:
\( \implies \frac{-5}{5} = -1 \)
Thus, the value is -1.
In simple words: The sum over product fraction simplifies directly to \( \frac{-5}{5} \), which gives -1.

Exam Tip: Writing reciprocal roots in sum-product form simplifies calculations and avoids finding individual roots.

Question 29. α, β are the zeroes of the quadratic polynomial x2 – (k+6)x +2 (2k – 1). Find the value of k if α + β = ½ α β (7)
Answer: For the polynomial \( x^2 - (k+6)x + 2(2k-1) \):
\( \alpha + \beta = k + 6 \)
\( \alpha\beta = 2(2k - 1) \)
We are given the condition:
\( \alpha + \beta = \frac{1}{2}\alpha\beta \)
Substituting the expressions:
\( \implies k + 6 = \frac{1}{2} [2(2k - 1)] \)
\( \implies k + 6 = 2k - 1 \)
\( \implies 2k - k = 6 + 1 \)
\( \implies k = 7 \)
Thus, the value of \( k \) is 7.
In simple words: Setting the sum \( k+6 \) equal to half of the product \( 2k-1 \) gives us a simple equation where \( k = 7 \).

Exam Tip: Simplify the factors on both sides (such as canceling out the \( \frac{1}{2} \) and 2) before expanding terms to save steps.

Question 30. if α, β are the zeroes of the quadratic polynomial x2 – 7x + 10, find the value of α3 + β3 (133)
Answer: For the polynomial \( x^2 - 7x + 10 \):
\( \alpha + \beta = 7 \)
\( \alpha\beta = 10 \)
We use the algebraic identity:
\( \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] \)
Substituting the values of the sum and product:
\( \implies \alpha^3 + \beta^3 = 7 \cdot [7^2 - 3(10)] \)
\( \implies 7 \cdot [49 - 30] \)
\( \implies 7 \cdot 19 = 133 \)
Thus, the value of the expression is 133.
In simple words: Rather than solving for the roots, we use the identity for the sum of cubes to compute the value directly as 133.

Exam Tip: Memorize the algebraic identity for \( \alpha^3 + \beta^3 \) as it avoids the need to find the individual roots.

Question 31. Find the sum and the product of the zeroes of cubic polynomial 2x3 -5x2 – 14x + 8 (5/2, -7, -4)
Answer: For a cubic polynomial of the form \( ax^3 + bx^2 + cx + d \):
- Sum of zeroes (\( \alpha + \beta + \gamma \)) = \( -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2} \)
- Sum of product of zeroes taken in pairs (\( \alpha\beta + \beta\gamma + \gamma\alpha \)) = \( \frac{c}{a} = \frac{-14}{2} = -7 \)
- Product of zeroes (\( \alpha\beta\gamma \)) = \( -\frac{d}{a} = -\frac{8}{2} = -4 \)
Thus, the values are \( \frac{5}{2} \), -7, and -4 respectively.
In simple words: The sum of the roots is \( \frac{5}{2} \), the sum of products in pairs is -7, and the product of the roots is -4.

Exam Tip: Be careful with signs when writing cubic coefficients; the formula for the sum of zeroes is \( -\frac{b}{a} \) while the product is \( -\frac{d}{a} \).

Question 32. Find the sum and product of the zeroes of quadratic polynomial x2 – 3
Answer: For the quadratic polynomial \( x^2 - 3 \), we can write it in standard form as \( x^2 + 0x - 3 \).
Here, \( a = 1 \), \( b = 0 \), and \( c = -3 \).
- Sum of zeroes = \( -\frac{b}{a} = -\frac{0}{1} = 0 \)
- Product of zeroes = \( \frac{c}{a} = \frac{-3}{1} = -3 \)
Thus, the sum is 0 and the product is -3.
In simple words: Since there is no \( x \) term, the roots must sum to 0. The constant term tells us that their product is -3.

Exam Tip: Write down the missing term with a coefficient of 0 to avoid choosing the wrong coefficients for the formula.

Question 33. If 1 is a zero of polynomial ax2 – 3(a-1) -1, then find the value of a (1)
Answer: Let \( p(x) = ax^2 - 3(a - 1)x - 1 \) (restoring the missing \( x \) variable in the polynomial expression):
Since 1 is a zero, we have \( p(1) = 0 \):
\( \implies a(1)^2 - 3(a - 1)(1) - 1 = 0 \)
\( \implies a - 3a + 3 - 1 = 0 \)
\( \implies -2a + 2 = 0 \)
\( \implies 2a = 2 \implies a = 1 \)
Thus, the value of \( a \) is 1.
In simple words: Plugging in \( x = 1 \) and setting the polynomial equal to zero lets us solve for \( a \), which gives \( a = 1 \).

Exam Tip: Simply substitute the given root into the polynomial to create an algebraic equation in terms of the unknown coefficient.

Question 34. If α, β are zeroes of quadratic polynomial x2 – (k + 6)x + 2(2k-1).Find k if α + β = 1/2αβ
Answer: For the given quadratic polynomial:
\( \alpha + \beta = k + 6 \)
\( \alpha\beta = 2(2k - 1) \)
We are given:
\( \alpha + \beta = \frac{1}{2}\alpha\beta \)
Substituting the sum and product:
\( \implies k + 6 = \frac{1}{2} [2(2k - 1)] \)
\( \implies k + 6 = 2k - 1 \)
\( \implies 2k - k = 6 + 1 \)
\( \implies k = 7 \)
Thus, the value of \( k \) is 7.
In simple words: Setting the sum \( k+6 \) equal to half of the product \( 2k-1 \) gives us a simple equation where \( k = 7 \).

Exam Tip: Verify your answer by plugging \( k=7 \) back into both sides of the condition to ensure they match.

Question 35. Divide (6 + 19x + x2 – 6x3) by (2 +5x – 3x2) and verify the division algorithm
Answer: Let us write both polynomials in standard decreasing order of powers:
Dividend \( p(x) = -6x^3 + x^2 + 19x + 6 \)
Divisor \( g(x) = -3x^2 + 5x + 2 \)
Performing the division:
- \( \frac{-6x^3}{-3x^2} = 2x \). Subtracting \( 2x(-3x^2 + 5x + 2) \):
\( \implies (-6x^3 + x^2 + 19x + 6) - (-6x^3 + 10x^2 + 4x) = -9x^2 + 15x + 6 \)
- \( \frac{-9x^2}{-3x^2} = 3 \). Subtracting \( 3(-3x^2 + 5x + 2) \):
\( \implies (-9x^2 + 15x + 6) - (-9x^2 + 15x + 6) = 0 \)
Thus, Quotient \( q(x) = 2x + 3 \) and Remainder \( r(x) = 0 \).
Verification of Division Algorithm:
\( p(x) = g(x) \cdot q(x) + r(x) \)
\( \implies (-3x^2 + 5x + 2)(2x + 3) + 0 \)
\( \implies -6x^3 - 9x^2 + 10x^2 + 15x + 4x + 6 \)
\( \implies -6x^3 + x^2 + 19x + 6 = p(x) \).
Hence, verified.
In simple words: We divide the standard form of the polynomial to find a quotient of \( 2x + 3 \) and remainder of 0. Multiplying the divisor by this quotient returns the original dividend, verifying the algorithm.

Exam Tip: Always rearrange polynomials in descending powers of \( x \) before performing division to keep calculations orderly.

Topic: Trigonometry

Question 1. If cotΘ = 15/8, evaluate (2 + 2sinΘ)(1 – sinΘ) / (1 + cosΘ)(2 – 2cosΘ) (225/64)
Answer: Let us simplify the given trigonometric expression:
\( \frac{(2 + 2\sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(2 - 2\cos\theta)} = \frac{2(1 + \sin\theta)(1 - \sin\theta)}{2(1 + \cos\theta)(1 - \cos\theta)} \)
Using the identity \( (1 + x)(1 - x) = 1 - x^2 \):
\( \implies \frac{1 - \sin^2\theta}{1 - \cos^2\theta} \)
Since \( 1 - \sin^2\theta = \cos^2\theta \) and \( 1 - \cos^2\theta = \sin^2\theta \):
\( \implies \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta \)
We are given \( \cot\theta = \frac{15}{8} \):
\( \implies \cot^2\theta = \left(\frac{15}{8}\right)^2 = \frac{225}{64} \)
Thus, the evaluated value is \( \frac{225}{64} \).
In simple words: Factoring out 2 and using basic identities simplifies our expression down to \( \cot^2\theta \). Squaring the given value of \( \frac{15}{8} \) gives the final answer of \( \frac{225}{64} \).

Exam Tip: Simplify the given expression using trigonometric identities before substituting the numerical ratios to avoid lengthy calculations.

Question 2. If 7 sin2 Ѳ + 3 cos2 Ѳ = 4, show that tanѲ= 1/√3
Answer: We are given:
\( 7\sin^2\theta + 3\cos^2\theta = 4 \)
We can write this as:
\( 4\sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = 4 \)
\( \implies 4\sin^2\theta + 3(\sin^2\theta + \cos^2\theta) = 4 \)
Since \( \sin^2\theta + \cos^2\theta = 1 \):
\( \implies 4\sin^2\theta + 3(1) = 4 \)
\( \implies 4\sin^2\theta = 4 - 3 \)
\( \implies 4\sin^2\theta = 1 \)
\( \implies \sin^2\theta = \frac{1}{4} \implies \sin\theta = \frac{1}{2} \)
Now we find \( \cos\theta \):
\( \cos^2\theta = 1 - \sin^2\theta = 1 - \frac{1}{4} = \frac{3}{4} \implies \cos\theta = \frac{\sqrt{3}}{2} \)
Now we calculate \( \tan\theta \):
\( \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \)
Hence, shown.
In simple words: Splitting the terms helps us use the identity \( \sin^2\theta + \cos^2\theta = 1 \) to find \( \sin\theta = \frac{1}{2} \). Calculating \( \cos\theta \) and dividing the two gives \( \tan\theta = \frac{1}{\sqrt{3}} \).

Exam Tip: Splitting \( 7\sin^2\theta \) into \( 4\sin^2\theta + 3\sin^2\theta \) is the cleanest way to introduce the fundamental identity \( \sin^2\theta + \cos^2\theta = 1 \).

Question 3. Evaluate: tan2 60˚ - 2 cos2 60˚ - ¾ sin2 45˚ - 4 sin2 30˚ (9/8)
Answer: Let us substitute the standard values:
\( \tan 60^\circ = \sqrt{3} \)
\( \cos 60^\circ = \frac{1}{2} \)
\( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
\( \sin 30^\circ = \frac{1}{2} \)
Now evaluate:
\( (\sqrt{3})^2 - 2 \left(\frac{1}{2}\right)^2 - \frac{3}{4} \left(\frac{1}{\sqrt{2}}\right)^2 - 4 \left(\frac{1}{2}\right)^2 \)
\( \implies 3 - 2\left(\frac{1}{4}\right) - \frac{3}{4}\left(\frac{1}{2}\right) - 4\left(\frac{1}{4}\right) \)
\( \implies 3 - \frac{1}{2} - \frac{3}{8} - 1 \)
\( \implies 2 - \frac{1}{2} - \frac{3}{8} \)
\( \implies \frac{16 - 4 - 3}{8} = \frac{9}{8} \)
Thus, the evaluated value is \( \frac{9}{8} \).
In simple words: We replace the trigonometric functions with their known numeric values and solve the basic fraction arithmetic to get \( \frac{9}{8} \).

Exam Tip: Memorize the standard trigonometric table thoroughly to avoid basic value errors during evaluations.

Question 4. Evaluate: sec2 54˚ - cot2 36˚ / Cosec2 57˚ - tan2 33˚ + 2 sin2 38˚ sec2 52˚ - sin2 45˚ (5/2)
Answer: Let us simplify the terms using complementary angle formulas:
1. \( \cot 36^\circ = \tan(90^\circ - 36^\circ) = \tan 54^\circ \)
\( \implies \sec^2 54^\circ - \cot^2 36^\circ = \sec^2 54^\circ - \tan^2 54^\circ = 1 \)
2. \( \tan 33^\circ = \cot(90^\circ - 33^\circ) = \cot 57^\circ \)
\( \implies \csc^2 57^\circ - \tan^2 33^\circ = \csc^2 57^\circ - \cot^2 57^\circ = 1 \)
Thus, the first fraction is \( \frac{1}{1} = 1 \).
3. For the term \( 2 \sin^2 38^\circ \sec^2 52^\circ \):
\( \sec 52^\circ = \csc(90^\circ - 52^\circ) = \csc 38^\circ \)
\( \implies 2 \sin^2 38^\circ \csc^2 38^\circ = 2 \left( \sin 38^\circ \times \frac{1}{\sin 38^\circ} \right)^2 = 2(1) = 2 \)
4. For the final term:
\( \sin^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \)
Now combine all evaluations:
\( \implies 1 + 2 - \frac{1}{2} = 3 - \frac{1}{2} = \frac{5}{2} \)
Thus, the evaluated value is \( \frac{5}{2} \).
In simple words: Using complementary angles lets us simplify the complex trigonometric terms to 1 and 2. Subtracting \( \frac{1}{2} \) gives the final result of \( \frac{5}{2} \).

Exam Tip: Identify complementary angle pairs (like 54 and 36, or 57 and 33) and convert one of them to simplify the expression.

Question 5. Evaluate: √2 tan2 45˚ + cos2 30˚ - sin2 60˚ (√2)
Answer: Let us substitute the standard values:
\( \tan 45^\circ = 1 \)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
Now evaluate:
\( \implies \sqrt{2}(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( \implies \sqrt{2} + \frac{3}{4} - \frac{3}{4} \)
\( \implies \sqrt{2} \)
Thus, the evaluated value is \( \sqrt{2} \).
In simple words: Since \( \cos 30^\circ \) and \( \sin 60^\circ \) have the same value, their squared terms cancel out, leaving only the first term, which is \( \sqrt{2} \).

Exam Tip: Notice when terms are equal and opposite so you can cancel them out directly without working out their numerical fractions.

Question 6. If sec2 Ѳ (1+sinѲ) (1-sinѲ) = k, find the value of k (k = 1)
Answer: Let us simplify the left-hand side of the equation:
\( \sec^2\theta (1 + \sin\theta)(1 - \sin\theta) = \sec^2\theta (1 - \sin^2\theta) \)
Since \( 1 - \sin^2\theta = \cos^2\theta \):
\( \implies \sec^2\theta \cdot \cos^2\theta \)
Since \( \sec\theta = \frac{1}{\cos\theta} \):
\( \implies \frac{1}{\cos^2\theta} \cdot \cos^2\theta = 1 \)
Thus, the value of \( k \) is 1.
In simple words: The terms simplify to \( \sec^2\theta \cdot \cos^2\theta \). Since they are reciprocals of each other, their product is 1.

Exam Tip: Always keep the fundamental identity \( \sin^2\theta + \cos^2\theta = 1 \) in mind as it is the most common simplification step.

Question 7. Evaluate: (sin 90˚ + cos 45˚ + cos 60˚)(cos 0˚ - sin 45˚ + sin 30˚) (7/4)
Answer: Substituting the standard values:
\( \sin 90^\circ = 1, \ \cos 45^\circ = \frac{1}{\sqrt{2}}, \ \cos 60^\circ = \frac{1}{2} \)
\( \cos 0^\circ = 1, \ \sin 45^\circ = \frac{1}{\sqrt{2}}, \ \sin 30^\circ = \frac{1}{2} \)
Now substitute into the expression:
\( \implies \left(1 + \frac{1}{\sqrt{2}} + \frac{1}{2}\right)\left(1 - \frac{1}{\sqrt{2}} + \frac{1}{2}\right) \)
\( \implies \left(\frac{3}{2} + \frac{1}{\sqrt{2}}\right)\left(\frac{3}{2} - \frac{1}{\sqrt{2}}\right) \)
Using the identity \( (a + b)(a - b) = a^2 - b^2 \):
\( \implies \left(\frac{3}{2}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 \)
\( \implies \frac{9}{4} - \frac{1}{2} \)
\( \implies \frac{9 - 2}{4} = \frac{7}{4} \)
Thus, the evaluated value is \( \frac{7}{4} \).
In simple words: We combine the numbers 1 and \( \frac{1}{2} \) to get \( \frac{3}{2} \). Applying the difference-of-squares formula leaves us with simple fraction arithmetic, resulting in \( \frac{7}{4} \).

Exam Tip: Grouping terms to match the form \( (a+b)(a-b) \) makes this complicated-looking product very easy to expand.

Question 8. Find the value of: 2 sin 68˚ / Cos 22˚ - 2 cot 15˚ / 5 tan 75˚ - 3 tan 45˚ tan 20˚ tan 40˚ tan 50˚ tan 70˚ / 5 (1)
Answer: Let us simplify each term using complementary angles:
- For the first term:
\( \cos 22^\circ = \sin(90^\circ - 22^\circ) = \sin 68^\circ \)
\( \implies \frac{2\sin 68^\circ}{\cos 22^\circ} = \frac{2\sin 68^\circ}{\sin 68^\circ} = 2 \)
- For the second term:
\( \tan 75^\circ = \cot(90^\circ - 75^\circ) = \cot 15^\circ \)
\( \implies \frac{2\cot 15^\circ}{5\tan 75^\circ} = \frac{2\cot 15^\circ}{5\cot 15^\circ} = \frac{2}{5} \)
- For the third term:
We know \( \tan 70^\circ = \cot 20^\circ \implies \tan 20^\circ \tan 70^\circ = 1 \)
And \( \tan 50^\circ = \cot 40^\circ \implies \tan 40^\circ \tan 50^\circ = 1 \)
Since \( \tan 45^\circ = 1 \):
\( \implies \frac{3 \tan 45^\circ \tan 20^\circ \tan 40^\circ \tan 50^\circ \tan 70^\circ}{5} = \frac{3(1)(1)(1)}{5} = \frac{3}{5} \)
Now combine the evaluated terms:
\( \implies 2 - \frac{2}{5} - \frac{3}{5} \)
\( \implies 2 - \left(\frac{2 + 3}{5}\right) = 2 - 1 = 1 \)
Thus, the final value is 1.
In simple words: Complementary relationships simplify the fraction terms to 2, \( \frac{2}{5} \), and \( \frac{3}{5} \). Subtracting them from 2 results in 1.

Exam Tip: Be sure to write down the intermediate step showing which angles are complementary to justify your cancellations.

Question 9. If sin (A + B) = 1, cos (A – B) = 1, find A and B (45˚, 45˚)
Answer: We are given:
\( \sin(A + B) = 1 \implies A + B = 90^\circ \) - (Equation 1)
\( \cos(A - B) = 1 \implies A - B = 0^\circ \implies A = B \) - (Equation 2)
Using Equation 2 in Equation 1:
\( \implies A + A = 90^\circ \)
\( \implies 2A = 90^\circ \implies A = 45^\circ \)
Since \( A = B \):
\( \implies B = 45^\circ \)
Thus, \( A = 45^\circ \) and \( B = 45^\circ \).
In simple words: Since \( \sin 90^\circ = 1 \) and \( \cos 0^\circ = 1 \), we get two simple linear equations that show both angles must be \( 45^\circ \).

Exam Tip: State the standard angles where sine is 1 and cosine is 1 to justify setting up the linear equations.

Question 10. If cos (40˚ + x) = sin 30˚, find the value of x (20˚)
Answer: We are given:
\( \cos(40^\circ + x) = \sin 30^\circ \)
Using the complementary angle relationship \( \sin\theta = \cos(90^\circ - \theta) \):
\( \implies \sin 30^\circ = \cos(90^\circ - 30^\circ) = \cos 60^\circ \)
Now substitute this back:
\( \implies \cos(40^\circ + x) = \cos 60^\circ \)
\( \implies 40^\circ + x = 60^\circ \)
\( \implies x = 20^\circ \)
Thus, the value of \( x \) is \( 20^\circ \).
In simple words: Since \( \sin 30^\circ \) is equal to \( \cos 60^\circ \), the angle inside the cosine on the left must be \( 60^\circ \), which gives \( x = 20^\circ \).

Exam Tip: Converting the right-hand side using complementary angles is more direct than substituting numerical values.

Question 11. If sin 2x = sin60˚ cos30˚ - cos60˚ sin30˚ , find x (15)
Answer: Let us substitute the standard values:
\( \sin 60^\circ = \frac{\sqrt{3}}{2}, \ \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( \cos 60^\circ = \frac{1}{2}, \ \sin 30^\circ = \frac{1}{2} \)
Now substitute into the right-hand side:
\( \sin 2x = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \)
\( \implies \sin 2x = \frac{3}{4} - \frac{1}{4} \)
\( \implies \sin 2x = \frac{2}{4} = \frac{1}{2} \)
Since \( \sin 30^\circ = \frac{1}{2} \):
\( \implies 2x = 30^\circ \implies x = 15^\circ \)
Thus, the value of \( x \) is \( 15^\circ \).
In simple words: The right-hand side simplifies to \( \frac{1}{2} \). Since the sine of \( 30^\circ \) is \( \frac{1}{2} \), we solve \( 2x = 30^\circ \) to find \( x = 15^\circ \).

Exam Tip: You can also use the trigonometric identity \( \sin(A - B) = \sin A\cos B - \cos A\sin B \) to simplify the RHS to \( \sin(60^\circ - 30^\circ) = \sin 30^\circ \) directly.

Question 12. Find the value of Ѳ in 2 cos 3Ѳ = 1 (20˚ )
Answer: We are given:
\( 2\cos 3\theta = 1 \)
\( \implies \cos 3\theta = \frac{1}{2} \)
Since \( \cos 60^\circ = \frac{1}{2} \):
\( \implies 3\theta = 60^\circ \)
\( \implies \theta = 20^\circ \)
Thus, the value of \( \theta \) is \( 20^\circ \).
In simple words: Dividing both sides by 2 gives \( \cos 3\theta = \frac{1}{2} \). Since cosine of \( 60^\circ \) is \( \frac{1}{2} \), the angle must be \( 20^\circ \).

Exam Tip: Be sure to solve for \( \theta \) by dividing the matched angle by 3 rather than multiplying.

Question 13. Sin 4A = cos (A - 20˚), where 4A is an acute angle, find the value of A (22˚)
Answer: We use the identity \( \sin\theta = \cos(90^\circ - \theta) \):
\( \implies \sin 4A = \cos(90^\circ - 4A) \)
Now substitute this into the given equation:
\( \implies \cos(90^\circ - 4A) = \cos(A - 20^\circ) \)
Comparing the angles:
\( \implies 90^\circ - 4A = A - 20^\circ \)
\( \implies 5A = 110^\circ \)
\( \implies A = 22^\circ \)
Thus, the value of \( A \) is \( 22^\circ \).
In simple words: Using complementary angles lets us set up a simple linear equation \( 90^\circ - 4A = A - 20^\circ \), which solves to give \( A = 22^\circ \).

Exam Tip: Clearly state the complementary angle conversion formula to show your working logic to the examiner.

Question 14. Find the acute angles A and B, A>B, if sin (A + 2B) = √3/2 and cos (A + 4B) = 0 (30˚, 15˚)
Answer: We are given:
\( \sin(A + 2B) = \frac{\sqrt{3}}{2} \implies A + 2B = 60^\circ \) - (Equation 1)
\( \cos(A + 4B) = 0 \implies A + 4B = 90^\circ \) - (Equation 2)
Subtracting Equation 1 from Equation 2:
\( \implies (A + 4B) - (A + 2B) = 90^\circ - 60^\circ \)
\( \implies 2B = 30^\circ \implies B = 15^\circ \)
Substitute \( B = 15^\circ \) into Equation 1:
\( \implies A + 2(15^\circ) = 60^\circ \)
\( \implies A + 30^\circ = 60^\circ \implies A = 30^\circ \)
Thus, the acute angles are \( A = 30^\circ \) and \( B = 15^\circ \).
In simple words: The sine and cosine equations give us two simple system equations in \( A \) and \( B \). Solving them gives \( A = 30^\circ \) and \( B = 15^\circ \).

Exam Tip: Check that your final values satisfy both conditions to ensure your solution is completely correct.

Question 15. Evaluate: sec (90 – Ѳ)cosecѲ – tan (90 – Ѳ)cotѲ + cos2 35 + cos2 55 / Tan5˚ tan15˚ tan45˚ tan75˚ tan85˚ (2)
Answer: Let us simplify the terms step-by-step:
1. For the first part:
\( \sec(90^\circ - \theta) = \csc\theta \)
\( \implies \sec(90^\circ - \theta)\csc\theta = \csc^2\theta \)
2. For the second part:
\( \tan(90^\circ - \theta) = \cot\theta \)
\( \implies \tan(90^\circ - \theta)\cot\theta = \cot^2\theta \)
Using the identity \( \csc^2\theta - \cot^2\theta = 1 \).
3. For the fraction term:
- Numerator: Since \( \cos 55^\circ = \sin 35^\circ \):
\( \cos^2 35^\circ + \cos^2 55^\circ = \cos^2 35^\circ + \sin^2 35^\circ = 1 \)
- Denominator: Using complementary pairs:
\( \tan 85^\circ = \cot 5^\circ \implies \tan 5^\circ \tan 85^\circ = 1 \)
\( \tan 75^\circ = \cot 15^\circ \implies \tan 15^\circ \tan 75^\circ = 1 \)
Since \( \tan 45^\circ = 1 \):
\( \implies \tan 5^\circ \tan 15^\circ \tan 45^\circ \tan 75^\circ \tan 85^\circ = 1 \times 1 \times 1 = 1 \)
So the fraction is \( \frac{1}{1} = 1 \).
Combining the parts:
\( \implies 1 + 1 = 2 \)
Thus, the evaluated value is 2.
In simple words: The first part simplifies to 1 using Pythagorean identities. The second part is a fraction that simplifies to 1, leading to a final result of 2.

Exam Tip: Grouping terms into complementary pairs makes complex trigonometric fractions very easy to resolve.

Question 16. If sinA – cosB = 0, prove that A + B = 90˚
Answer: We are given:
\( \sin A - \cos B = 0 \)
\( \implies \sin A = \cos B \)
Using the complementary identity \( \cos B = \sin(90^\circ - B) \):
\( \implies \sin A = \sin(90^\circ - B) \)
Comparing the angles:
\( \implies A = 90^\circ - B \)
\( \implies A + B = 90^\circ \)
Hence, proven.
In simple words: Since sine and cosine values are equal for complementary angles, \( A \) and \( B \) must add up to \( 90^\circ \).

Exam Tip: Showing the step \( \sin A = \cos B \) is crucial for establishing the complementary relationship between the angles.

Question 17. If sinѲ + cosѲ / sinѲ – cosѲ = 5/3 , evaluate 7 tanѲ + 2 / 2tanѲ + 7 (2)
Answer: Let us divide the numerator and denominator of the left-hand side of the given equation by \( \cos\theta \):
\( \frac{\frac{\sin\theta}{\cos\theta} + 1}{\frac{\sin\theta}{\cos\theta} - 1} = \frac{5}{3} \)
\( \implies \frac{\tan\theta + 1}{\tan\theta - 1} = \frac{5}{3} \)
By cross-multiplication:
\( \implies 3(\tan\theta + 1) = 5(\tan\theta - 1) \)
\( \implies 3\tan\theta + 3 = 5\tan\theta - 5 \)
\( \implies 2\tan\theta = 8 \implies \tan\theta = 4 \)
Now substitute \( \tan\theta = 4 \) into the expression we need to evaluate:
\( \implies \frac{7\tan\theta + 2}{2\tan\theta + 7} = \frac{7(4) + 2}{2(4) + 7} = \frac{28 + 2}{8 + 7} = \frac{30}{15} = 2 \)
Thus, the evaluated value is 2.
In simple words: Dividing the first equation by \( \cos\theta \) lets us find \( \tan\theta = 4 \). Plugging this value into our target expression gives the result of 2.

Exam Tip: Dividing throughout by \( \cos\theta \) is a helpful technique to convert sine-cosine equations directly into tangent equations.

Question 18. What is the maximum value of 1/secѲ
Answer: We know that:
\( \frac{1}{\sec\theta} = \cos\theta \)
The maximum value of the cosine function \( \cos\theta \) for any real angle \( \theta \) is 1.
Thus, the maximum value of \( \frac{1}{\sec\theta} \) is 1.
In simple words: Since the expression is equal to \( \cos\theta \), its highest possible value is 1.

Exam Tip: Always convert complex reciprocal ratios to their basic trigonometric counterparts (like cosine or sine) to analyze their range.

Question 19. If A, B and C are interior angles of triangle ABC, show that cos [B+C]/2 = sin A/2
Answer: Since \( A \), \( B \), and \( C \) are interior angles of a triangle \( ABC \):
\( A + B + C = 180^\circ \)
\( \implies B + C = 180^\circ - A \)
Dividing both sides of the equation by 2:
\( \implies \frac{B + C}{2} = 90^\circ - \frac{A}{2} \)
Taking the cosine function on both sides:
\( \implies \cos\left(\frac{B + C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) \)
Since \( \cos(90^\circ - \theta) = \sin\theta \):
\( \implies \cos\left(\frac{B + C}{2}\right) = \sin\left(\frac{A}{2}\right) \)
Hence, shown.
In simple words: Using the fact that all angles in a triangle add to \( 180^\circ \), we rearrange terms and apply complementary identities to prove the relation.

Exam Tip: State the angle sum property of triangles clearly as the starting point of your proof.

Question 20. If x = a sinѲ, y = b tanѲ. Prove that a2/x2 - b2/y2 = 1
Answer: We are given:
\( x = a\sin\theta \implies \frac{a}{x} = \frac{1}{\sin\theta} = \csc\theta \)
\( y = b\tan\theta \implies \frac{b}{y} = \frac{1}{\tan\theta} = \cot\theta \)
Substituting these into the left-hand side of the expression:
\( \implies \frac{a^2}{x^2} - \frac{b^2}{y^2} = \csc^2\theta - \cot^2\theta \)
Using the standard identity \( \csc^2\theta - \cot^2\theta = 1 \):
\( \implies 1 \)
Hence, proven.
In simple words: Expressing \( \frac{a}{x} \) and \( \frac{b}{y} \) in terms of trigonometric ratios allows us to use the Pythagorean identity to prove the equation equals 1.

Exam Tip: Expressing the variable fractions in terms of cosecant and cotangent is the most elegant way to solve this identity.

Question 21. Prove that: 1 / 1 + sinѲ + 1 / 1 – sinѲ = 2 sec2 Ѳ
Answer: Let us simplify the left-hand side of the equation:
\( \text{LHS} = \frac{1}{1 + \sin\theta} + \frac{1}{1 - \sin\theta} \)
Taking the common denominator:
\( \implies \frac{(1 - \sin\theta) + (1 + \sin\theta)}{(1 + \sin\theta)(1 - \sin\theta)} \)
\( \implies \frac{2}{1 - \sin^2\theta} \)
Since \( 1 - \sin^2\theta = \cos^2\theta \):
\( \implies \frac{2}{\cos^2\theta} = 2\sec^2\theta = \text{RHS} \)
Hence, proven.
In simple words: Adding the fractions simplifies the numerator to 2 and the denominator to \( \cos^2\theta \), which gives \( 2\sec^2\theta \).

Exam Tip: Make sure you show the expansion of the denominator as \( 1 - \sin^2\theta \) clearly before converting it to \( \cos^2\theta \).

Question 22. Prove that: sinѲ / 1 + cosѲ + 1 + cosѲ / sinѲ = 2cosecѲ
Answer: Let us simplify the left-hand side of the equation:
\( \text{LHS} = \frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} \)
Taking the common denominator:
\( \implies \frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)} \)
\( \implies \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1 + \cos\theta)} \)
Since \( \sin^2\theta + \cos^2\theta = 1 \):
\( \implies \frac{1 + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)} \)
\( \implies \frac{2 + 2\cos\theta}{\sin\theta(1 + \cos\theta)} \)
\( \implies \frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} \)
Canceling out \( (1 + \cos\theta) \) from both numerator and denominator:
\( \implies \frac{2}{\sin\theta} = 2\csc\theta = \text{RHS} \)
Hence, proven.
In simple words: We combine the fractions and use \( \sin^2\theta + \cos^2\theta = 1 \) to simplify the numerator to \( 2(1+\cos\theta) \). Canceling terms leaves \( \frac{2}{\sin\theta} \), which is \( 2\csc\theta \).

Exam Tip: Grouping \( \sin^2\theta \) and \( \cos^2\theta \) together in the numerator helps make the identity substitution obvious to the examiner.

Question 23. Prove: √ (1 + sin A / 1 – sin A) = cos A / 1 – sin A
Answer: Let us simplify the left-hand side:
\( \text{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}} \)
Rationalize the denominator inside the square root by multiplying by \( \frac{1 + \sin A}{1 + \sin A} \):
\( \implies \sqrt{\frac{(1 + \sin A)^2}{(1 - \sin A)(1 + \sin A)}} \)
\( \implies \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} \)
Since \( 1 - \sin^2 A = \cos^2 A \):
\( \implies \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} \)
Now, let us multiply both numerator and denominator of this result by \( 1 - \sin A \):
\( \implies \frac{(1 + \sin A)(1 - \sin A)}{\cos A(1 - \sin A)} \)
\( \implies \frac{1 - \sin^2 A}{\cos A(1 - \sin A)} \)
\( \implies \frac{\cos^2 A}{\cos A(1 - \sin A)} \)
\( \implies \frac{\cos A}{1 - \sin A} = \text{RHS} \)
Hence, proven.
In simple words: Rationalizing the square root helps simplify the expression to \( \frac{1+\sin A}{\cos A} \). Multiplying top and bottom by \( 1-\sin A \) yields the target form \( \frac{\cos A}{1-\sin A} \).

Exam Tip: Multiplying top and bottom by the conjugate of the denominator is a standard technique to clear radicals in trigonometric proofs.

Question 24. Prove that sin (90 – Ѳ) cos (90 – Ѳ) = tanѲ / 1 + tan2 Ѳ
Answer: Let us simplify the left-hand side using complementary angle identities:
\( \text{LHS} = \sin(90^\circ - \theta)\cos(90^\circ - \theta) = \cos\theta\sin\theta \)
Now, let us simplify the right-hand side:
\( \text{RHS} = \frac{\tan\theta}{1 + \tan^2\theta} \)
Since \( 1 + \tan^2\theta = \sec^2\theta \):
\( \implies \frac{\tan\theta}{\sec^2\theta} \)
Expressing in terms of sine and cosine:
\( \implies \frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}} \)
\( \implies \frac{\sin\theta}{\cos\theta} \cdot \cos^2\theta = \sin\theta\cos\theta \)
Since \( \text{LHS} = \text{RHS} \), the identity is verified.
In simple words: The left side simplifies to \( \sin\theta\cos\theta \). Converting the right side to sine and cosine terms yields the exact same result, proving the identity.

Exam Tip: Simplifying LHS and RHS separately to the same expression is a very reliable strategy for trigonometric proofs.