CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set 04

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Short Answer Type Questions

Question. Prove that: \( (\sqrt{3} + 1)(3 - \cot 30^\circ) = \tan^3 60^\circ - 2 \sin 60^\circ \)
Answer: Sol. L.H.S. \( = (\sqrt{3} + 1)(3 - \cot 30^\circ) = (\sqrt{3} + 1)(3 - \sqrt{3}) \)
...[ \( \because \cot 30^\circ = \sqrt{3} \)
\( = 3\sqrt{3} - \sqrt{3} \cdot \sqrt{3} + 3 - \sqrt{3} \)
\( = 3\sqrt{3} - 3 + 3 - \sqrt{3} = 2\sqrt{3} \)
Also, R.H.S. \( = \tan^3 60^\circ - 2 \sin 60^\circ \)
\( = (\sqrt{3})^3 - 2 \frac{\sqrt{3}}{2} = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3} \)
\( \therefore \) L.H.S. = R.H.S. ...(Hence proved)

Question. If \( \sqrt{3} \tan \theta = 1 \), then find the value of \( \sin^2 \theta - \cos^2 \theta \).
Answer: Sol. Given: \( \sqrt{3} \tan \theta = 1 \)
\( \Rightarrow \tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \tan \theta = \tan 30^\circ \)
\( \Rightarrow \theta = 30^\circ \)
\( \therefore \sin^2 \theta - \cos^2 \theta = \sin^2 30^\circ - \cos^2 30^\circ \)
\( = \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} - \frac{3}{4} = \frac{1-3}{4} = \frac{-2}{4} = -\frac{1}{2} \)

PRACTICE QUESTIONS
CLASS X : CHAPTER - 8
INTRODUCTION TO TRIGONOMETRY
TRIGONOMETRIC IDENTITIES

1. Prove that \( \frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} = \sin \theta + \cos \theta \).

Answer:
1. Since 1 - tanθ = \( \frac{cosθ - sinθ}{cosθ} \), the first term is \( \frac{cos²θ}{cosθ - sinθ} \).
2. Since 1 - cotθ = \( \frac{sinθ - cosθ}{sinθ} \), the second term is \( \frac{sin²θ}{sinθ - cosθ} \).
3. Sum = \( \frac{cos²θ - sin²θ}{cosθ - sinθ} \) = (cosθ - sinθ)\( \frac{cosθ + sinθ}{cosθ - sinθ} \) = cosθ + sinθ.
4. Hence, LHS = RHS.

Teacher's Note:
a) Clear the small fractions by multiplying the top and bottom by cosθ or sinθ.
b) The difference of two squares cancels the common factor, so state that cosθ is not equal to sinθ.
c) Do not cancel any factor before both terms are combined.

2. Prove that \( \frac{1}{2} \left\{ \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} \right\} = \frac{1}{\sin \theta} \).

Answer:
1. Common denominator of the bracket is sinθ(1 + cosθ), so its numerator is sin²θ + (1 + cosθ)².
2. Expand: sin²θ + 1 + 2cosθ + cos²θ = 2 + 2cosθ = 2(1 + cosθ).
3. The bracket equals 2\( \frac{1 + cosθ}{sinθ(1 + cosθ)} \) = \( \frac{2}{sinθ.} \)
4. Multiply by \( \frac{1}{2} \): LHS = \( \frac{1}{sinθ} \) = RHS. Hence proved.

Teacher's Note:
a) The identity sin²θ + cos²θ = 1 turns the numerator into a multiple of (1 + cosθ).
b) The outer \( \frac{1}{2} \) cancels the 2 that appears after expansion.
c) Assume sinθ is not zero and cosθ is not minus one.

3. Prove that: \( \frac{\tan^{3} \alpha}{1 + \tan^{2} \alpha} + \frac{\cot^{3} \alpha}{1 + \cot^{2} \alpha} = \sec \alpha \cosec \alpha - 2 \sin \alpha \cos \alpha \).

Answer:
1. Since 1 + tan²α = sec²α, the first term is tan³α cos²α = \( \frac{sin³α}{cosα.} \)
2. Since 1 + cot²α = cosec²α, the second term is cot³α sin²α = \( \frac{cos³α}{sinα.} \)
3. Sum = \( \frac{sin³α}{cosα} \) + \( \frac{cos³α}{sinα} \) = \( \frac{sin⁴α + cos⁴α}{sinα cosα} \).
4. Since sin⁴α + cos⁴α = (sin²α + cos²α)² - 2sin²α cos²α = 1 - 2sin²α cos²α, the sum = \( \frac{1}{sinα cosα} \) - 2sinα cosα.
5. \( \frac{1}{sinα cosα} \) = secα cosecα, so the sum = secα cosecα - 2sinα cosα. Hence proved.

Teacher's Note:
a) Replace 1 + tan²α by sec²α and 1 + cot²α by cosec²α first.
b) Students often forget that sin⁴α + cos⁴α needs the squared identity.
c) Splitting the sum over sinα cosα gives the product form directly.

4. Prove that: \( \frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A} = 1 + \tan A + \cot A = 1 + \sec A \cosec A \).

Answer:
1. \( \frac{tanA}{1 - cotA} \) = \( \frac{sin²A}{cosA(sinA - cosA)} \), after multiplying the top and bottom by sinA.
2. \( \frac{cotA}{1 - tanA} \) = \( \frac{cos²A}{sinA(cosA - sinA)} \), after multiplying the top and bottom by cosA.
3. Sum = \( \frac{sin³A - cos³A}{sinA cosA (sinA - cosA)} \).
4. Since sin³A - cos³A = (sinA - cosA)(1 + sinA cosA), the sum = \( \frac{1 + sinA cosA}{sinA cosA} \) = 1 + \( \frac{1}{sinA cosA} \).
5. Also 1 + tanA + cotA = 1 + \( \frac{sin²A + cos²A}{sinA cosA} \) = 1 + \( \frac{1}{sinA cosA} \).
6. Since \( \frac{1}{sinA cosA} \) = secA cosecA, every expression is equal. Hence proved.

Teacher's Note:
a) Each term has a denominator of the form 1 - (ratio), so multiply by sinA or cosA to clear it.
b) The factor a³ - b³ = (a - b)(a² + ab + b²) is the key step; look for it.
c) Three expressions are linked here, and the middle one is the one to prove.

5. Prove that: \( \frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \sqrt{\frac{1 - \cos A}{1 + \cos A}} \).

Answer:
1. Use the half angle forms: 1 - cosA = 2sin²(\( \frac{A}{2} \)), 1 + cosA = 2cos²(\( \frac{A}{2} \)) and sinA = 2sin(\( \frac{A}{2} \))cos(\( \frac{A}{2} \)).
2. Numerator = 2sin²(\( \frac{A}{2} \)) + 2sin(\( \frac{A}{2} \))cos(\( \frac{A}{2} \)) = 2sin(\( \frac{A}{2} \))[sin(\( \frac{A}{2} \)) + cos(\( \frac{A}{2} \))].
3. Denominator = 2cos²(\( \frac{A}{2} \)) + 2sin(\( \frac{A}{2} \))cos(\( \frac{A}{2} \)) = 2cos(\( \frac{A}{2} \))[cos(\( \frac{A}{2} \)) + sin(\( \frac{A}{2} \))].
4. LHS = sin\( \frac{\( \frac{A}{2} \)}{cos} \)(\( \frac{A}{2} \)) = tan(\( \frac{A}{2} \)).
5. RHS = \( \sqrt{\frac{2\sin^{2}(A/2)}{2\cos^{2}(A/2)}} \) = tan(\( \frac{A}{2} \)), for 0 < A < 90°.
6. Hence, LHS = RHS.

Teacher's Note:
a) Half angle formulas turn the 1 ± cosA terms into squares.
b) The common factor sin(\( \frac{A}{2} \)) + cos(\( \frac{A}{2} \)) cancels in the numerator and denominator.
c) The square root gives the positive value tan(\( \frac{A}{2} \)), so take 0 < A < 90°.

6. Prove that \( (\tan A + \cosec B)^{2} - (\cot B - \sec A)^{2} = 2 \tan A \cot B (\cosec A + \sec B) \).

Answer:
1. Expand LHS: tan²A + 2tanA cosecB + cosec²B - cot²B + 2cotB secA - sec²A.
2. Group the terms: (tan²A - sec²A) + (cosec²B - cot²B) = -1 + 1 = 0.
3. So LHS = 2tanA cosecB + 2cotB secA = 2\( \frac{sinA + cosB}{sinB cosA} \).
4. RHS = 2tanA cotB cosecA + 2tanA cotB secB = \( \frac{2cosB}{cosA sinB} \) + \( \frac{2sinA}{cosA sinB} \) = 2\( \frac{sinA + cosB}{sinB cosA} \).
5. Hence, LHS = RHS.

Teacher's Note:
a) The squared terms pair up into the standard identities sec²A - tan²A = 1 and cosec²B - cot²B = 1.
b) Write every term in sines and cosines to compare the two sides.
c) Check the sign of each cross term, because the second bracket is subtracted.

7. Prove that: \( \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A \).

Answer:
1. Common denominator is cosA(1 + sinA), so the numerator is cos²A + (1 + sinA)².
2. Expand: cos²A + 1 + 2sinA + sin²A = 2 + 2sinA = 2(1 + sinA).
3. The sum = 2\( \frac{1 + sinA}{cosA(1 + sinA)} \) = \( \frac{2}{cosA} \) = 2secA. Hence proved.

Teacher's Note:
a) This is the same pattern as Question 2, so the method carries over.
b) The numerator collapses to 2(1 + sinA) only after using sin²A + cos²A = 1.

8. Prove that: \( \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \cosec A + \cot A \).

Answer:
1. Multiply the top and bottom by (cosA + sinA + 1).
2. Denominator = (cosA + sinA)² - 1 = 2sinA cosA.
3. Numerator = (cosA + 1)² - sin²A = cos²A + 2cosA + 1 - sin²A = 2cosA(1 + cosA).
4. LHS = 2cosA\( \frac{1 + cosA}{2sinA cosA} \) = \( \frac{1 + cosA}{sinA} \) = \( \frac{1}{sinA} \) + \( \frac{cosA}{sinA} \) = cosecA + cotA. Hence proved.

Teacher's Note:
a) Multiplying by the conjugate-type factor (cosA + sinA + 1) removes the awkward denominator.
b) Use the identity sin²A = 1 - cos²A to simplify the numerator.
c) Cancelling the common factor cosA is allowed once it has been shown to be a factor of both parts.

9. Prove that: \( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{\sin^{2} A - \cos^{2} A} = \frac{2}{2\sin^{2} A - 1} = \frac{2}{1 - 2\cos^{2} A} \).

Answer:
1. Sum = [(sinA + cosA)² + (sinA - cosA)²]/[(sinA - cosA)(sinA + cosA)].
2. Numerator = 2sin²A + 2cos²A = 2.
3. Denominator = sin²A - cos²A.
4. So the sum = \( \frac{2}{sin²A - cos²A} \).
5. Since cos²A = 1 - sin²A, sin²A - cos²A = 2sin²A - 1 = 1 - 2cos²A. Hence all three forms are equal.

Teacher's Note:
a) The numerator of the sum is found by adding two squares; the cross terms cancel.
b) Replacing cos²A by 1 - sin²A gives the second and third forms.
c) Keep the denominator as sin²A - cos²A until the final step.

10. Prove that: \( \frac{\sin A}{\cot A + \cosec A} = 2 + \frac{\sin A}{\cot A - \cosec A} \).

Answer:
1. cotA + cosecA = \( \frac{cosA + 1}{sinA} \), so the LHS = \( \frac{sin²A}{1 + cosA} \).
2. cotA - cosecA = \( \frac{cosA - 1}{sinA} \), so \( \frac{sinA}{cotA - cosecA} \) = \( \frac{sin²A}{cosA - 1} \).
3. Since sin²A = 1 - cos²A = (1 - cosA)(1 + cosA), \( \frac{sin²A}{cosA - 1} \) = -(1 + cosA).
4. RHS = 2 - (1 + cosA) = 1 - cosA.
5. LHS = \( \frac{sin²A}{1 + cosA} \) = (1 - cosA)\( \frac{1 + cosA}{1 + cosA} \) = 1 - cosA. So LHS = RHS.

Teacher's Note:
a) Convert cot and cosec into sines and cosines with the common denominator sinA.
b) The sign of cosA - 1 is negative, so keep the minus sign when factorising.
c) Both sides simplify to 1 - cosA, which is the best check.

11. Prove that: \( \sqrt{\frac{\sec A - 1}{\sec A + 1}} + \sqrt{\frac{\sec A + 1}{\sec A - 1}} = 2 \cosec A \).

Answer:
1. secA - 1 = \( \frac{1 - cosA}{cosA} \) and secA + 1 = \( \frac{1 + cosA}{cosA} \), so \( \frac{secA - 1}{secA + 1} \) = \( \frac{1 - cosA}{1 + cosA} \).
2. Multiply the top and bottom by (1 - cosA): \( \frac{1 - cosA}{1 + cosA} \) = (1 - cosA)\( \frac{²}{sin²A.} \)
3. So the first term = \( \frac{1 - cosA}{sinA} \), taking 0 < A < 90°.
4. Similarly, the second term = \( \frac{1 + cosA}{sinA.} \)
5. Sum = \( \frac{1 - cosA + 1 + cosA}{sinA} \) = \( \frac{2}{sinA} \) = 2cosecA. Hence proved.

Teacher's Note:
a) Write the secant terms as cosine fractions before taking any square root.
b) The square root of a squared expression is positive only when the expression is positive, so state the range of A.
c) The two terms are reciprocal-type expressions, and their sum collapses neatly.

12. Prove that: \( \frac{1}{\cosec A - \cot A} - \frac{1}{\sin A} = \frac{1}{\sin A} - \frac{1}{\cosec A + \cot A} \).

Answer:
1. Multiply top and bottom of \( \frac{1}{cosecA - cotA} \) by (cosecA + cotA): the denominator becomes cosec²A - cot²A = 1.
2. So \( \frac{1}{cosecA - cotA} \) = cosecA + cotA = \( \frac{1 + cosA}{sinA.} \)
3. Similarly, \( \frac{1}{cosecA + cotA} \) = cosecA - cotA = \( \frac{1 - cosA}{sinA.} \)
4. LHS = \( \frac{1 + cosA}{sinA} \) - \( \frac{1}{sinA} \) = \( \frac{cosA}{sinA.} \)
5. RHS = \( \frac{1}{sinA} \) - \( \frac{1 - cosA}{sinA} \) = \( \frac{cosA}{sinA.} \) So LHS = RHS.

Teacher's Note:
a) The product (cosecA - cotA)(cosecA + cotA) equals cosec²A - cot²A, which is 1.
b) This gives a reciprocal rule: \( \frac{1}{cosecA - cotA} \) = cosecA + cotA.
c) Both sides come out as cotA, so check the final value as well as the form.

13. Prove that: \( \frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} = \sec \theta + \tan \theta = \frac{1 + \sin \theta}{\cos \theta} \).

Answer:
1. Since sec²θ - tan²θ = 1, (secθ - tanθ)(secθ + tanθ) = 1.
2. Let S = secθ + tanθ. Then secθ - tanθ = \( \frac{1}{S.} \)
3. Numerator = S - 1. Denominator = 1 - (secθ - tanθ) = 1 - \( \frac{1}{S} \) = \( \frac{S - 1}{S.} \)
4. LHS = (S - 1) × \( \frac{S}{S - 1} \) = S, provided S is not equal to 1.
5. So LHS = secθ + tanθ = \( \frac{1 + sinθ}{cosθ} \), since \( \frac{1 + sinθ}{cosθ} \) = \( \frac{1}{cosθ} \) + \( \frac{sinθ}{cosθ.} \) Hence proved.

Teacher's Note:
a) Naming the sum S = secθ + tanθ makes the fraction easy to see.
b) The reciprocal property of sec²θ - tan²θ = 1 does most of the work.
c) S cannot be 1, because that would make the denominator zero.

14. If x = a sinθ + b cosθ and y = a cosθ + b sinθ, prove that x² + y² = a² + b².

Answer:
1. The printed value of y has a plus sign, but with y = a cosθ + b sinθ, x² + y² = a² + b² + 4ab sinθ cosθ, which is not equal to a² + b². The correct relation is y = a cosθ - b sinθ, and that is used below.
2. x² = a²sin²θ + 2ab sinθ cosθ + b²cos²θ.
3. y² = a²cos²θ - 2ab sinθ cosθ + b²sin²θ.
4. Add: x² + y² = a²(sin²θ + cos²θ) + b²(cos²θ + sin²θ) = a² + b².

Teacher's Note:
a) The cross terms 2ab sinθ cosθ and -2ab sinθ cosθ cancel only when the signs of y are opposite.
b) The printed y has a plus sign, which is a misprint; with the minus sign the identity holds.
c) Squaring and adding two rotated expressions gives a constant, which is a standard pattern.

15. If \( \sec \theta + \tan \theta = m \), show that \( \left( \frac{m^{2} - 1}{m^{2} + 1} \right) = \sin \theta \).

Answer:
1. Since sec²θ - tan²θ = 1, (secθ - tanθ) = \( \frac{1}{m.} \)
2. Adding the two equations: 2secθ = m + \( \frac{1}{m} \) = \( \frac{m² + 1}{m} \), so secθ = \( \frac{m² + 1}{2m} \).
3. Subtracting: 2tanθ = m - \( \frac{1}{m} \) = \( \frac{m² - 1}{m} \), so tanθ = \( \frac{m² - 1}{2m} \).
4. sinθ = \( \frac{tanθ}{secθ} \) = \( \frac{m² - 1}{m² + 1} \).
5. Hence, \( \frac{m^{2} - 1}{m^{2} + 1} = \sin \theta \).

Teacher's Note:
a) Find secθ and tanθ separately by adding and subtracting the two relations.
b) sinθ = \( \frac{tanθ}{secθ} \) is the quickest link to sine.
c) The reciprocal m is never zero here, so the division in Step 2 is valid.

16. Prove that: \( \frac{1 + \cos \theta + \sin \theta}{1 + \cos \theta - \sin \theta} = \frac{1 + \sin \theta}{\cos \theta} \).

Answer:
1. Cross-multiply: cosθ(1 + cosθ + sinθ) = (1 + sinθ)(1 + cosθ - sinθ).
2. LHS = cosθ + cos²θ + sinθ cosθ.
3. RHS = 1 + cosθ - sinθ + sinθ + sinθ cosθ - sin²θ = 1 + cosθ + sinθ cosθ - sin²θ.
4. Equate: cosθ + cos²θ = 1 + cosθ - sin²θ, so cos²θ + sin²θ = 1, which is true. Hence proved.

Teacher's Note:
a) Cross-multiplication avoids messy fractions on both sides.
b) Most of the terms cancel, leaving only sin²θ + cos²θ = 1.
c) Expand (1 + sinθ)(1 + cosθ - sinθ) with care, since the minus sign on sinθ changes the sign of sin²θ.

17. Prove that \( \sec^{4} A (1 - \sin^{4} A) - 2 \tan^{2} A = 1 \).

Answer:
1. 1 - sin⁴A = (1 - sin²A)(1 + sin²A) = cos²A(1 + sin²A).
2. sec⁴A × cos²A(1 + sin²A) = \( \frac{1 + sin²A}{cos²A} \) = sec²A + \( \frac{sin²A}{cos²A.} \)
3. sec²A + tan²A = (1 + tan²A) + tan²A = 1 + 2tan²A.
4. So LHS = 1 + 2tan²A - 2tan²A = 1. Hence proved.

Teacher's Note:
a) Factor 1 - sin⁴A as a difference of squares to get cos²A.
b) Writing \( \frac{sin²A}{cos²A} \) as tan²A links the expression to the given term.
c) The final cancellation is exact, so the value 1 is the answer.

18. If \( \cosec \theta - \sin \theta = m \) and \( \sec \theta - \cos \theta = n \), prove that \( (m^{2} n)^{2/3} + (m n^{2})^{2/3} = 1 \).

Answer:
1. cosecθ - sinθ = \( \frac{1 - sin²θ}{sinθ} \) = \( \frac{cos²θ}{sinθ} \), so m = \( \frac{cos²θ}{sinθ.} \)
2. secθ - cosθ = \( \frac{sin²θ}{cosθ} \), so n = \( \frac{sin²θ}{cosθ.} \)
3. m²n = (\( \frac{cos⁴θ}{sin²θ} \))(\( \frac{sin²θ}{cosθ} \)) = cos³θ, so \( (m²n)^{\frac{2}{3}} \) = cos²θ.
4. mn² = (\( \frac{cos²θ}{sinθ} \))(\( \frac{sin⁴θ}{cos²θ} \)) = sin³θ, so \( (mn²)^{\frac{2}{3}} \) = sin²θ.
5. Sum = cos²θ + sin²θ = 1. Hence proved.

Teacher's Note:
a) Convert both given relations to sines and cosines first.
b) The cube inside each bracket cancels exactly after the power \( \frac{2}{3} \) is applied.
c) Watch the powers of m and n: m²n gives cos³θ and mn² gives sin³θ.

19. If \( \tan \theta + \sin \theta = m \) and \( \tan \theta - \sin \theta = n \), show that \( m^{2} - n^{2} = 4\sqrt{mn} \).

Answer:
1. m² - n² = (m - n)(m + n) = (2sinθ)(2tanθ) = 4sinθ tanθ.
2. mn = (tanθ + sinθ)(tanθ - sinθ) = tan²θ - sin²θ = tan²θ(1 - cos²θ) = tan²θ sin²θ.
3. So √(mn) = tanθ sinθ, taking the positive root.
4. Hence m² - n² = 4 √(mn). Hence proved.

Teacher's Note:
a) The difference of squares gives the product (m - n)(m + n) at once.
b) tan²θ - sin²θ factorises as tan²θ sin²θ using 1 - cos²θ = sin²θ.
c) The square root is valid only when tanθ sinθ is positive, so take 0 < θ < 90°.

20. If \( a \cos \theta - b \sin \theta = c \), prove that \( (a \sin \theta + b \cos \theta) = \pm \sqrt{a^{2} + b^{2} - c^{2}} \).

Answer:
1. Let P = a sinθ + b cosθ. Then P² = a²sin²θ + 2ab sinθ cosθ + b²cos²θ.
2. Also c² = (a cosθ - b sinθ)² = a²cos²θ - 2ab sinθ cosθ + b²sin²θ.
3. Add: P² + c² = a²(sin²θ + cos²θ) + b²(cos²θ + sin²θ) = a² + b².
4. So P² = a² + b² - c², and P = ±√(a² + b² - c²). Hence proved.

Teacher's Note:
a) The cross terms cancel when the squares are added, so no value of θ is needed.
b) The sign ± appears because only P² is fixed by the data.
c) This is a common board pattern: square both the required expression and the given one.

21. If \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \), prove that \( \cos \theta - \sin \theta = \sqrt{2} \sin \theta \).

Answer:
1. From the given relation, sinθ = (√2 - 1)cosθ.
2. cosθ - sinθ = cosθ - (√2 - 1)cosθ = (2 - √2)cosθ.
3. √2 sinθ = √2(√2 - 1)cosθ = (2 - √2)cosθ.
4. So cosθ - sinθ = √2 sinθ. Hence proved.

Teacher's Note:
a) Express everything in terms of cosθ, since cosθ is not zero here.
b) Multiplying out √2(√2 - 1) gives 2 - √2, which matches the other side.
c) Keep the surds exact; decimals can hide the equality.

22. If \( \left( \frac{x}{a} \sin \theta - \frac{y}{b} \cos \theta \right) = 1 \) and \( \left( \frac{x}{a} \cos \theta + \frac{y}{b} \sin \theta \right) = 1 \), prove that \( \left( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \right) = 2 \).

Answer:
1. Let p = \( \frac{x}{a} \) and q = \( \frac{y}{b.} \) The two equations are p sinθ - q cosθ = 1 and p cosθ + q sinθ = 1.
2. Square the first: p²sin²θ - 2pq sinθ cosθ + q²cos²θ = 1.
3. Square the second: p²cos²θ + 2pq sinθ cosθ + q²sin²θ = 1.
4. Add the two squared equations: p²(sin²θ + cos²θ) + q²(cos²θ + sin²θ) = 2.
5. So p² + q² = 2, that is, \( \frac{x²}{a²} \) + \( \frac{y²}{b²} \) = 2. Hence proved.

Teacher's Note:
a) Substitute p = \( \frac{x}{a} \) and q = \( \frac{y}{b} \) to keep the algebra short.
b) The cross terms cancel when the two squares are added.
c) Notice that adding squares of the two rotated forms always gives the sum of the squared coefficients.

23. If \( (\tan \theta + \sin \theta) = m \) and \( (\tan \theta - \sin \theta) = n \), prove that \( (m^{2} - n^{2})^{2} = 16 mn \).

Answer:
1. m² - n² = (m - n)(m + n) = (2sinθ)(2tanθ) = 4sinθ tanθ.
2. (m² - n²)² = 16 sin²θ tan²θ.
3. mn = tan²θ - sin²θ = tan²θ sin²θ, as in Question 19.
4. So 16mn = 16 tan²θ sin²θ = (m² - n²)². Hence proved.

Teacher's Note:
a) Squaring removes the square root that appears in Question 19.
b) Reuse the earlier result that mn equals tan²θ sin²θ.
c) The factor 16 comes from squaring the factor 4 in m² - n².

24. If \( \cosec \theta - \sin \theta = a^{3} \) and \( \sec \theta - \cos \theta = b^{3} \), prove that \( a^{2} b^{2} (a^{2} + b^{2}) = 1 \).

Answer:
1. As in Question 18: cosecθ - sinθ = \( \frac{cos²θ}{sinθ} \) and secθ - cosθ = \( \frac{sin²θ}{cosθ.} \)
2. a³ = \( \frac{cos²θ}{sinθ} \) and b³ = \( \frac{sin²θ}{cosθ.} \)
3. a²b² = \( (a³)^{\frac{2}{3}} \) \( (b³)^{\frac{2}{3}} \) = \( (cos²θ/sinθ)^{\frac{2}{3}} \)\( (sin²θ/cosθ)^{\frac{2}{3}} \) = \( cos^{\frac{2}{3}} \)θ \( sin^{\frac{2}{3}} \)θ.
4. a² = \( cos^{\frac{4}{3}} \)θ \( sin^{\frac{-2}{3}} \)θ and b² = \( sin^{\frac{4}{3}} \)θ \( cos^{\frac{-2}{3}} \)θ.
5. So a²b²(a² + b²) = \( cos^{\frac{2}{3}} \)θ \( sin^{\frac{2}{3}} \)θ × (\( cos^{\frac{4}{3}} \)θ \( sin^{\frac{-2}{3}} \)θ + \( sin^{\frac{4}{3}} \)θ \( cos^{\frac{-2}{3}} \)θ) = cos²θ + sin²θ = 1. Hence proved.

Teacher's Note:
a) Take the cube root of each given equation to find a² and b² in terms of sines and cosines.
b) Use exponent rules carefully; the \( \frac{2}{3} \) power is applied to each factor separately.
c) The final cancellation gives sin²θ + cos²θ, so the answer is 1.

25. If \( a \cos^{3} \theta + 3a \sin^{2} \theta \cos \theta = m \) and \( a \sin^{3} \theta + 3a \sin \theta \cos^{2} \theta = n \), prove that \( (m + n)^{2/3} + (m - n)^{2/3} = 2a^{2/3} \).

Answer:
1. m + n = a(cos³θ + 3sin²θ cosθ + sin³θ + 3sinθ cos²θ) = a(cosθ + sinθ)³.
2. m - n = a(cos³θ + 3sin²θ cosθ - sin³θ - 3sinθ cos²θ) = a(cosθ - sinθ)³.
3. \( (m + n)^{\frac{2}{3}} \) = \( a^{\frac{2}{3}} \)(cosθ + sinθ)², since \( (a × cube)^{\frac{2}{3}} \) = \( a^{\frac{2}{3}} \) × square.
4. \( (m - n)^{\frac{2}{3}} \) = \( a^{\frac{2}{3}} \)(cosθ - sinθ)².
5. Sum = \( a^{\frac{2}{3}} \)[(cosθ + sinθ)² + (cosθ - sinθ)²] = \( a^{\frac{2}{3}} \) × 2(sin²θ + cos²θ) = 2\( a^{\frac{2}{3}} \). Hence proved.

Teacher's Note:
a) Recognise the binomial expansions of (cosθ + sinθ)³ and (cosθ - sinθ)³.
b) Take the \( \frac{2}{3} \) power only after the cube is identified.
c) The two squares add to 2, which gives the factor 2 on the right side.

26. Prove that \( \sqrt{\sec^{2} \theta + \cosec^{2} \theta} = \tan \theta + \cot \theta \).

Answer:
1. sec²θ + cosec²θ = \( \frac{1}{cos²θ} \) + \( \frac{1}{sin²θ} \) = \( \frac{sin²θ + cos²θ}{sin²θ cos²θ} \) = \( \frac{1}{sin²θ cos²θ} \).
2. So the LHS = \( \frac{1}{sinθ cosθ} \), taking sinθ cosθ > 0.
3. tanθ + cotθ = \( \frac{sinθ}{cosθ} \) + \( \frac{cosθ}{sinθ} \) = \( \frac{sin²θ + cos²θ}{sinθ cosθ} \) = \( \frac{1}{sinθ cosθ} \).
4. Hence LHS = RHS, for 0 < θ < 90°. Hence proved.

Teacher's Note:
a) Add the two fractions over the common denominator sin²θ cos²θ.
b) The square root returns the absolute value, so the sign of sinθ cosθ must be stated.
c) Both sides give the same reciprocal product, which confirms the identity.

27. Prove the identity: \( \frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta} \).

Answer:
1. RHS = \( \frac{1}{\( \frac{1 - sinθ}{cosθ} \)} \) = \( \frac{cosθ}{1 - sinθ} \).
2. Cross-multiply: (sinθ - cosθ + 1)(1 - sinθ) = cosθ(sinθ + cosθ - 1).
3. Expand the left: sinθ - sin²θ - cosθ + sinθcosθ + 1 - sinθ = 1 - sin²θ - cosθ + sinθ cosθ = cos²θ - cosθ + sinθ cosθ.
4. Expand the right: sinθ cosθ + cos²θ - cosθ.
5. Both sides are equal, so LHS = RHS. Hence proved.

Teacher's Note:
a) Convert secθ - tanθ into a single fraction before taking the reciprocal.
b) Cross-multiplying is safe because both denominators are non-zero.
c) Watch the sign of each term when expanding (sinθ - cosθ + 1)(1 - sinθ).

28. Prove the identity: \( \sec^{6} \theta = \tan^{6} \theta + 3 \tan^{2} \theta \sec^{2} \theta + 1 \).

Answer:
1. Let t = tan²θ. Then sec²θ = 1 + t, so sec⁶θ = (1 + t)³.
2. Expand: (1 + t)³ = 1 + 3t + 3t² + t³.
3. RHS = t³ + 3t(1 + t) + 1 = t³ + 3t + 3t² + 1.
4. Both are equal to 1 + 3t + 3t² + t³. Hence proved.

Teacher's Note:
a) Substituting t = tan²θ turns the identity into an ordinary cube expansion.
b) Do not forget that sec⁶θ means (sec²θ)³.
c) The identity is a special case of (1 + t)³, so it is good practice for binomial expansion.

29. Prove the identity: \( (\sin A + \cosec A)^{2} + (\cos A + \sec A)^{2} = 7 + \tan^{2} A + \cot^{2} A \).

Answer:
1. Expand the first bracket: sin²A + 2 + cosec²A.
2. Expand the second bracket: cos²A + 2 + sec²A.
3. Add: (sin²A + cos²A) + 4 + cosec²A + sec²A = 5 + cosec²A + sec²A.
4. Use cosec²A = 1 + cot²A and sec²A = 1 + tan²A: sum = 5 + 1 + cot²A + 1 + tan²A = 7 + tan²A + cot²A. Hence proved.

Teacher's Note:
a) The cross terms 2sinA cosecA and 2cosA secA each equal 2.
b) Replace cosec²A and sec²A by 1 + cot²A and 1 + tan²A.
c) Count the constants carefully: 2 + 2 from the cross terms, and 1 + 1 + 1 from the identities.

30. If \( x \sin^{3} \theta + y \cos^{3} \theta = \sin \theta \cos \theta \) and \( x \sin \theta = y \cos \theta \), prove that \( x^{2} + y^{2} = 1 \).

Answer:
1. Write x = t cosθ and y = t sinθ, where t = \( \frac{x}{cosθ} \) = \( \frac{y}{sinθ.} \) This satisfies x sinθ = y cosθ.
2. Substitute into the first equation: t cosθ sin³θ + t sinθ cos³θ = sinθ cosθ.
3. Factor: t sinθ cosθ (sin²θ + cos²θ) = sinθ cosθ, so t sinθ cosθ = sinθ cosθ.
4. Since sinθ cosθ is not zero, t = 1. So x = cosθ and y = sinθ.
5. x² + y² = cos²θ + sin²θ = 1. Hence proved.

Teacher's Note:
a) The condition x sinθ = y cosθ means (x, y) lies on the line through the origin at angle θ.
b) Write the common ratio t explicitly; it removes the unknowns.
c) Dividing by sinθ cosθ is valid only when neither sinθ nor cosθ is zero.

31. If \( \sec \theta = x + \frac{1}{4x} \), prove that \( \sec \theta + \tan \theta = 2x \) or \( \frac{1}{2x} \).

Answer:
1. tan²θ = sec²θ - 1 = x² + \( \frac{1}{2} \) + \( \frac{1}{16x²} \) - 1 = x² - \( \frac{1}{2} \) + \( \frac{1}{16x²} \).
2. This is a perfect square: tan²θ = (x - \( \frac{1}{4x} \))².
3. So tanθ = x - \( \frac{1}{4x} \) or tanθ = -(x - \( \frac{1}{4x} \)).
4. In the first case, secθ + tanθ = (x + \( \frac{1}{4x} \)) + (x - \( \frac{1}{4x} \)) = 2x.
5. In the second case, secθ + tanθ = (x + \( \frac{1}{4x} \)) - (x - \( \frac{1}{4x} \)) = \( \frac{1}{2x} \). Hence proved.

Teacher's Note:
a) Use sec²θ - 1 = tan²θ to find tan²θ before taking any root.
b) The expression under the root is a perfect square, which is why two answers appear.
c) The two cases correspond to the two signs of tanθ.

32. Prove that \( \left( 1 + \frac{1}{\tan^{2} A} \right) \left( 1 + \frac{1}{\cot^{2} A} \right) = \frac{1}{\sin^{2} A - \sin^{4} A} \).

Answer:
1. 1 + \( \frac{1}{tan²A} \) = 1 + cot²A = cosec²A.
2. 1 + \( \frac{1}{cot²A} \) = 1 + tan²A = sec²A.
3. LHS = cosec²A sec²A = \( \frac{1}{sin²A cos²A} \).
4. RHS: sin²A - sin⁴A = sin²A(1 - sin²A) = sin²A cos²A, so RHS = \( \frac{1}{sin²A cos²A} \).
5. LHS = RHS. Hence proved.

Teacher's Note:
a) Reciprocal squares turn 1 + \( \frac{1}{tan²A} \) into 1 + cot²A.
b) Factor sin²A out of sin²A - sin⁴A to see cos²A.
c) Both sides equal the same product, so the check is quick.

33. If \( \cot \theta + \tan \theta = x \) and \( \sec \theta - \cos \theta = y \), prove that \( (x^{2} y)^{2/3} - (x y^{2})^{2/3} = 1 \).

Answer:
1. cotθ + tanθ = \( \frac{cos²θ + sin²θ}{sinθ cosθ} \) = \( \frac{1}{sinθ cosθ} \), so x = \( \frac{1}{sinθ cosθ} \).
2. secθ - cosθ = \( \frac{1 - cos²θ}{cosθ} \) = \( \frac{sin²θ}{cosθ} \), so y = \( \frac{sin²θ}{cosθ.} \)
3. x²y = (\( \frac{1}{sin²θ cos²θ} \)) × (\( \frac{sin²θ}{cosθ} \)) = \( \frac{1}{cos³θ} \), so \( (x²y)^{\frac{2}{3}} \) = \( \frac{1}{cos²θ.} \)
4. xy² = (\( \frac{1}{sinθ cosθ} \)) × (\( \frac{sin⁴θ}{cos²θ} \)) = \( \frac{sin³θ}{cos³θ} \), so \( (xy²)^{\frac{2}{3}} \) = \( \frac{sin²θ}{cos²θ.} \)
5. Difference = \( \frac{1 - sin²θ}{cos²θ} \) = \( \frac{cos²θ}{cos²θ} \) = 1. Hence proved.

Teacher's Note:
a) Simplify x and y to single trigonometric terms before applying the \( \frac{2}{3} \) power.
b) The cube root inside \( (x²y)^{\frac{2}{3}} \) cancels the cube in the denominator.
c) The final step uses 1 - sin²θ = cos²θ.

34. If \( \frac{\cos \alpha}{\cos \beta} = m \) and \( \frac{\cos \alpha}{\sin \beta} = n \), show that \( (m^{2} + n^{2}) \cos^{2} \beta = n^{2} \).

Answer:
1. m² + n² = cos²α (\( \frac{1}{cos²β} \) + \( \frac{1}{sin²β} \)).
2. Multiply by cos²β: (m² + n²)cos²β = cos²α (1 + \( \frac{cos²β}{sin²β} \)).
3. 1 + \( \frac{cos²β}{sin²β} \) = \( \frac{sin²β + cos²β}{sin²β} \) = \( \frac{1}{sin²β.} \)
4. So (m² + n²)cos²β = \( \frac{cos²α}{sin²β.} \)
5. n² = \( \frac{cos²α}{sin²β} \), so (m² + n²)cos²β = n². Hence shown.

Teacher's Note:
a) Factor cos²α out of m² + n² first.
b) The term 1 + cot²β appears here; it equals cosec²β.
c) Compare the final expression with n² = \( \frac{cos²α}{sin²β.} \)

35. If \( \cosec \theta - \sin \theta = a \) and \( \sec \theta - \cos \theta = b \), prove that \( a^{2} b^{2} (a^{2} + b^{2} + 3) = 1 \).

Answer:
1. a = \( \frac{cos²θ}{sinθ} \) and b = \( \frac{sin²θ}{cosθ.} \)
2. a²b² = (\( \frac{cos⁴θ}{sin²θ} \))(\( \frac{sin⁴θ}{cos²θ} \)) = cos²θ sin²θ.
3. a² + b² = \( \frac{cos⁶θ + sin⁶θ}{sin²θ cos²θ} \).
4. So a²b²(a² + b²) = cos⁶θ + sin⁶θ.
5. cos⁶θ + sin⁶θ = (sin²θ + cos²θ)³ - 3sin²θ cos²θ(sin²θ + cos²θ) = 1 - 3sin²θ cos²θ.
6. a²b² × 3 = 3sin²θ cos²θ, so a²b²(a² + b² + 3) = 1 - 3sin²θ cos²θ + 3sin²θ cos²θ = 1. Hence proved.

Teacher's Note:
a) Express a and b in sines and cosines first, as in Question 18.
b) The sum of sixth powers is written as a cube of (sin²θ + cos²θ) minus a correction term.
c) The terms -3sin²θ cos²θ and +3sin²θ cos²θ cancel exactly.

36. If \( x = r \sin A \cos C \), \( y = r \sin A \sin C \) and \( z = r \cos A \), prove that \( r^{2} = x^{2} + y^{2} + z^{2} \).

Answer:
1. x² + y² = r²sin²A cos²C + r²sin²A sin²C = r²sin²A(cos²C + sin²C) = r²sin²A.
2. z² = r²cos²A.
3. x² + y² + z² = r²sin²A + r²cos²A = r²(sin²A + cos²A) = r².
4. Hence r² = x² + y² + z². Hence proved.

Teacher's Note:
a) Add the squares of x and y first, because the C terms combine.
b) Factor r² out of every term to see the pattern clearly.
c) This is the squared distance formula in spherical coordinates.

37. If \( \tan A = n \tan B \) and \( \sin A = m \sin B \), prove that \( \cos^{2} A = \frac{m^{2} - 1}{n^{2} - 1} \).

Answer:
1. Let S = sin²B. Then sin²A = m²S, and tan²A = n²tan²B.
2. tan²A = \( \frac{sin²A}{1 - sin²A} \) and tan²B = \( \frac{S}{1 - S} \), so \( \frac{m²S}{1 - m²S} \) = \( \frac{n²S}{1 - S} \).
3. Cancel S (S is not zero): m²(1 - S) = n²(1 - m²S).
4. Expand: m² - m²S = n² - n²m²S, so S m²(n² - 1) = n² - m².
5. So S = \( \frac{n² - m²}{m²(n² - 1)} \).
6. cos²A = 1 - sin²A = 1 - m²S = 1 - \( \frac{n² - m²}{n² - 1} \) = \( \frac{m² - 1}{n² - 1} \). Hence proved.

Teacher's Note:
a) Introduce S = sin²B to reduce the number of unknowns.
b) Cancelling S is valid only when sinB is not zero.
c) Keep the signs of n² - m² and m² - 1 distinct; a sign slip changes the final form.
d) Rewriting 1 - \( \frac{n² - m²}{n² - 1} \) gives the answer in one line.

38. If \( \sin \theta + \sin^{2} \theta = 1 \), find the value of \( \cos^{12} \theta + 3\cos^{10} \theta + 3\cos^{8} \theta + \cos^{6} \theta + 2\cos^{4} \theta + 2\cos^{2} \theta - 2 \).

Answer:
1. From sinθ + sin²θ = 1, sinθ = 1 - sin²θ = cos²θ.
2. Let c = cos²θ. Then sinθ = c, so sin²θ = 1 - c. Also c² = cos⁴θ = sin²θ = 1 - c.
3. So c² + c = 1, that is, c(c + 1) = 1.
4. The expression = c⁶ + 3c⁵ + 3c⁴ + c³ + 2c² + 2c - 2 = c³(c + 1)³ + 2c(c + 1) - 2.
5. Substitute c(c + 1) = 1: expression = 1 + 2 - 2 = 1.
6. Hence the value is 1.

Teacher's Note:
a) Write cos²θ as a single variable c, which makes the powers manageable.
b) The grouping c³(c + 1)³ = (c(c + 1))³ is the key step.
c) Note that cos⁴θ = sin²θ, which gives the relation c² + c = 1.

39. Prove that: \( (1 - \sin \theta + \cos \theta)^{2} = 2(1 + \cos \theta)(1 - \sin \theta) \).

Answer:
1. LHS = 1 + sin²θ + cos²θ - 2sinθ + 2cosθ - 2sinθ cosθ.
2. Using sin²θ + cos²θ = 1: LHS = 2 - 2sinθ + 2cosθ - 2sinθ cosθ.
3. RHS = 2(1 - sinθ + cosθ - sinθ cosθ) = 2 - 2sinθ + 2cosθ - 2sinθ cosθ.
4. LHS = RHS. Hence proved.

Teacher's Note:
a) Square the trinomial term by term, and remember the double products.
b) The two cross terms with sinθ and cosθ match the expansion of the right side.
c) Students often miss the factor -2sinθ cosθ when squaring.

40. If \( \sin \theta + \sin^{2} \theta = 1 \), prove that \( \cos^{2} \theta + \cos^{4} \theta = 1 \).

Answer:
1. sinθ = 1 - sin²θ = cos²θ.
2. cos⁴θ = (cos²θ)² = sin²θ.
3. cos²θ + cos⁴θ = sinθ + sin²θ = 1. Hence proved.

Teacher's Note:
a) The given relation is exactly sin θ = cos²θ in disguise.
b) Squaring cos²θ gives cos⁴θ, which equals sin²θ.
c) This is the same key relation used in Question 38.

41. If \( a \sec \theta + b \tan \theta + c = 0 \) and \( p \sec \theta + q \tan \theta + r = 0 \), prove that \( (br - qc)^{2} - (pc - ar)^{2} = (aq - bp)^{2} \).

Answer:
1. Let S = secθ and T = tanθ. Then aS + bT = -c and pS + qT = -r.
2. Let D = aq - bp. Solving the linear pair gives S = \( \frac{br - qc}{D} \) and T = \( \frac{pc - ar}{D.} \)
3. Check: aS + bT = [a(br - qc) + b(pc - ar)]/D = \( \frac{bpc - aqc}{D} \) = -c, as required.
4. Since S² - T² = 1, (br - qc)² - (pc - ar)² = D²(S² - T²) = D².
5. So (br - qc)² - (pc - ar)² = (aq - bp)². Hence proved.

Teacher's Note:
a) Treat secθ and tanθ as two unknowns in a pair of linear equations.
b) The determinant D = aq - bp appears in both solutions.
c) The identity sec²θ - tan²θ = 1 is what converts the answer into a clean square.

42. If \( \sin \theta + \sin^{2} \theta + \sin^{3} \theta = 1 \), then prove that \( \cos^{6} \theta - 4\cos^{4} \theta + 8\cos^{2} \theta = 4 \).

Answer:
1. Let s = sinθ, so s + s² + s³ = 1 and cos²θ = 1 - s².
2. Put u = cos²θ. The LHS is u³ - 4u² + 8u, and u³ - 4u² + 8u - 4 = 1 - 3s² - s⁴ - s⁶.
3. From the given relation, s³ = 1 - s - s², so s⁴ = s·s³ = 2s - 1.
4. s⁶ = (s³)² = (1 - s - s²)² = 1 - 2s - s² + s⁴ + 2s³ = 2 - 2s - 3s².
5. So 1 - 3s² - s⁴ - s⁶ = 1 - 3s² - (2s - 1) - (2 - 2s - 3s²) = 0.
6. Therefore u³ - 4u² + 8u = 4, that is, cos⁶θ - 4cos⁴θ + 8cos²θ = 4. Hence proved.

Teacher's Note:
a) Reduce high powers of sinθ using the given equation, one power at a time.
b) Squaring s³ gives s⁶, which is a common place to slip.
c) Expanding (1 - s² )³ and the other powers carefully is the longest part of the proof.

43. If \( \tan^{2} \theta = 1 - a^{2} \), prove that \( \sec \theta + \tan^{3} \theta \cosec \theta = (2 - a^{2})^{3/2} \).

Answer:
1. tan³θ cosecθ = tan²θ × (\( \frac{tanθ}{sinθ} \)) = tan²θ × secθ.
2. So LHS = secθ + tan²θ secθ = secθ(1 + tan²θ) = sec³θ.
3. sec²θ = 1 + tan²θ = 1 + (1 - a²) = 2 - a².
4. So sec³θ = \( (sec²θ)^{\frac{3}{2}} \) = \( (2 - a²)^{\frac{3}{2}} \), taking secθ positive. Hence proved.

Teacher's Note:
a) Write tan³θ cosecθ as tan²θ times tanθ cosecθ, and tanθ cosecθ = secθ.
b) Factor secθ out of the sum to get sec³θ.
c) The power \( \frac{3}{2} \) is the same as taking the square root and then cubing; keep the sign of secθ positive.

44. If \( x = a \sec \theta + b \tan \theta \) and \( y = a \tan \theta + b \sec \theta \), prove that \( x^{2} - y^{2} = a^{2} - b^{2} \).

Answer:
1. x² = a²sec²θ + 2ab secθ tanθ + b²tan²θ.
2. y² = a²tan²θ + 2ab tanθ secθ + b²sec²θ.
3. x² - y² = a²(sec²θ - tan²θ) + b²(tan²θ - sec²θ) = a² - b², since the cross terms cancel and sec²θ - tan²θ = 1.
4. Hence x² - y² = a² - b². Hence proved.

Teacher's Note:
a) The cross terms 2ab secθ tanθ cancel when the squares are subtracted.
b) Group the coefficients of a² and b² separately.
c) Recognise sec²θ - tan²θ = 1 as the key identity.

45. If \( 3 \sin \theta + 5 \cos \theta = 5 \), prove that \( 5 \sin \theta - 3 \cos \theta = \pm 3 \).

Answer:
1. (3sinθ + 5cosθ)² + (5sinθ - 3cosθ)² = 9sin²θ + 30sinθ cosθ + 25cos²θ + 25sin²θ - 30sinθ cosθ + 9cos²θ.
2. The cross terms cancel, so the sum = 34(sin²θ + cos²θ) = 34.
3. Since 3sinθ + 5cosθ = 5, its square is 25.
4. So (5sinθ - 3cosθ)² = 34 - 25 = 9.
5. Therefore 5sinθ - 3cosθ = ±3. Hence proved.

Teacher's Note:
a) Adding the squares of a linear form and its perpendicular form gives a constant.
b) The 34 comes from 3² + 5² = 9 + 25.
c) The ± sign is needed because only the square of the second expression is found.

Question. Show that \( \tan^4 \theta + \tan^2 \theta = \sec^4 \theta - \sec^2 \theta \).
Answer: Sol. L.H.S. \( \tan^4 \theta + \tan^2 \theta = \tan^2 \theta (\tan^2 \theta + 1) \)
\( = \tan^2 \theta \cdot \sec^2 \theta \) ...[\( \because \sec^2 \theta = \tan^2 \theta + 1 \)]
\( = (\sec^2 \theta - 1) \sec^2 \theta \) ...[\( \because \tan^2 \theta = \sec^2 \theta - 1 \)]
\( = \sec^4 \theta - \sec^2 \theta = \text{R.H.S.} \) ...(Hence proved)

Long Answer Type Questions

Question. If \( \text{cosec } \theta + \cot \theta = P \), then prove that \( \cos \theta = \frac{P^2 - 1}{P^2 + 1} \).
Answer: Sol. R.H.S \( = \frac{P^2 - 1}{P^2 + 1} \)
\( = \frac{(\text{cosec } \theta + \cot \theta)^2 - 1}{(\text{cosec } \theta + \cot \theta)^2 + 1} \)
\( = \frac{\text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta + \cot^2 \theta - 1}{\text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta + \cot^2 \theta + 1} \)
\( = \frac{\cot^2 \theta + \cot^2 \theta + 2 \text{cosec } \theta \cot \theta}{\text{cosec}^2 \theta + \text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta} \) ...[\( \text{cosec}^2 \theta - 1 = \cot^2 \theta \)]
\( = \frac{2 \cot^2 \theta + 2 \text{cosec } \theta \cot \theta}{2 \text{cosec}^2 \theta + 2 \text{cosec } \theta \cot \theta} \)
\( = \frac{2 \cot \theta (\cot \theta + \text{cosec } \theta)}{2 \text{cosec } \theta (\text{cosec } \theta + \cot \theta)} = \frac{\cot \theta}{\text{cosec } \theta} \)
\( = \frac{\cos \theta}{\sin \theta} \times \frac{\sin \theta}{1} = \cos \theta = \text{L.H.S.} \)

Question. If \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \), then prove that \( \tan \theta = 1 \) or \( \frac{1}{2} \).
Answer: Sol. \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \) ...[Given
Dividing both sides by \( \sin^2 \theta \), we get
\( \frac{1}{\sin^2 \theta} + 1 = 3 \cdot \frac{\cos \theta}{\sin \theta} \)
\( \Rightarrow \text{cosec}^2 \theta + 1 = 3 \cdot \cot \theta \)
\( \Rightarrow 1 + \cot^2 \theta + 1 = 3 \cot \theta \) ...[\( \because \text{cosec}^2 \theta - \cot^2 \theta = 1 \)]
\( \Rightarrow \cot^2 \theta - 3 \cot \theta + 2 = 0 \)
\( \Rightarrow \cot^2 \theta - 2 \cot \theta - \cot \theta + 2 = 0 \)
\( \Rightarrow \cot \theta(\cot \theta - 2) - 1(\cot \theta - 2) = 0 \)
\( \Rightarrow (\cot \theta - 2)(\cot \theta - 1) = 0 \)
\( \cot \theta - 1 = 0 \text{ or } \cot \theta - 2 = 0 \)
\( \Rightarrow \cot \theta = 1 \text{ or } \cot \theta = 2 \)
\( \therefore \tan \theta = 1 \text{ or } \frac{1}{2} \) ...[\( \because \tan \theta = \frac{1}{\cot \theta} \)]
...(Hence proved)

Question. If \( \tan \theta + \sec \theta = l \), then prove that \( \sec \theta = \frac{l^2 + 1}{2l} \).
Answer: Sol. Given: \( \tan \theta + \sec \theta = l \) ...(i)
Multiplying the numerator and denominator of L.H.S. of above equation by \( (\sec \theta - \tan \theta) \), we get,
\( (\tan \theta + \sec \theta) \times \frac{\sec \theta - \tan \theta}{\sec \theta - \tan \theta} = l \)
\( \Rightarrow \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} = l \) ...[\( (a+b)(a-b) = a^2 - b^2 \)]
\( \Rightarrow \frac{1}{\sec \theta - \tan \theta} = l \) ...[\( \sec^2 \theta - \tan^2 \theta = 1 \)]
\( \Rightarrow \sec \theta - \tan \theta = \frac{1}{l} \) ...(ii)
Adding equations (i) and (ii), we get
\( 2 \sec \theta = l + \frac{1}{l} \)
\( \Rightarrow \sec \theta = \frac{l^2 + 1}{2l} \) ...(Hence Proved)

Question. If \( \sin \theta + \cos \theta = p \) and \( \sec \theta + \text{cosec } \theta = q \), then prove that \( q(p^2 - 1) = 2p \).
Answer: Sol. Given: \( \sin \theta + \cos \theta = p \) ...(i)
and \( \sec \theta + \text{cosec } \theta = q \)
\( \Rightarrow \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = q \)
\( \Rightarrow \frac{\sin \theta + \cos \theta}{\sin \theta \cdot \cos \theta} = q \)
\( \Rightarrow \frac{p}{\sin \theta \cdot \cos \theta} = q \) ...[From (i)]
\( \Rightarrow \sin \theta \cdot \cos \theta = \frac{p}{q} \) ...(ii)
Squaring both sides of (i), we get
\( (\sin \theta + \cos \theta)^2 = p^2 \)
\( \Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cdot \cos \theta = p^2 \) ...[\( (a+b)^2 = a^2 + 2ab + b^2 \)]
\( \Rightarrow 1 + 2 \sin \theta \cdot \cos \theta = p^2 \) ...[\( \sin^2 \theta + \cos^2 \theta = 1 \)]
\( \Rightarrow 1 + 2 \cdot \frac{p}{q} = p^2 \) ...[From (ii)]
\( \Rightarrow \frac{q + 2p}{q} = p^2 \Rightarrow q + 2p = p^2q \)
\( \Rightarrow 2p = p^2q - q \)
\( \Rightarrow q(p^2 - 1) = 2p \) ...(Hence proved)

TRIGONOMETRIC RATIO (T.R.) OF SOME SPECIFIC ANGLES

The angles \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ \) are angles for which we have values of T.R.

• \( \sin \theta \uparrow \) when \( \theta \uparrow, 0 \le \theta \le 90^\circ \)
• \( \cos \theta \downarrow \) when \( \theta \uparrow, 0 \le \theta \le 90^\circ \)
• \( \tan \theta, \cot \theta \) are not defined for \( \theta = 90^\circ \) & \( 0^\circ \) respectively.
• \( \csc \theta, \sec \theta \) are not defined when \( \theta = 0^\circ \) & \( 90^\circ \) respectively.
• \( \sin \theta = \cos \theta \) for only \( \theta = 45^\circ \)
• \( 180^\circ = \pi^c \)
• \( 30^\circ = \left( \frac{\pi}{6} \right)^c \); \( 45^\circ = \left( \frac{\pi}{4} \right)^c \); \( 60^\circ = \left( \frac{\pi}{3} \right)^c \); \( 90^\circ = \left( \frac{\pi}{2} \right)^c \)

EXAMPLES

Question. Evaluate each of the following in the simplest form :
(i) \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \)
(ii) \( \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \)
Answer: (i) \( \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ \)
\( = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \)
(ii) \( \sin 60^\circ \cos 45^\circ + \cos 60^\circ \sin 45^\circ \)
\( = \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \times \frac{1}{\sqrt{2}} = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)

Question. Evaluate the following expression :
(i) \( \tan 60^\circ \csc^2 45^\circ + \sec^2 60^\circ \tan 45^\circ \)
(ii) \( 4\cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ \)
Answer: (i) \( \tan 60^\circ \csc^2 45^\circ + \sec^2 60^\circ \tan 45^\circ \)
\( = \sqrt{3} \times (\sqrt{2})^2 + (2)^2 \times 1 = \sqrt{3} \times 2 + 4 = 4 + 2\sqrt{3} \)
(ii) \( 4\cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + \cos^2 90^\circ \)
\( = 4(1)^2 - (2)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 + 0 = 4 - 4 + \frac{3}{4} + 0 = \frac{3}{4} \)

Question. Show that :
(i) \( 2(\cos^2 45^\circ + \tan^2 60^\circ) - 6(\sin^2 45^\circ - \tan^2 30^\circ) = 6 \)
(ii) \( 2(\cos^4 60^\circ + \sin^4 30^\circ) - (\tan^2 60^\circ + \cot^2 45^\circ) + 3\sec^2 30^\circ = \frac{1}{4} \)
Answer: (i) \( 2(\cos^2 45^\circ + \tan^2 60^\circ) - 6(\sin^2 45^\circ - \tan^2 30^\circ) \)
\( = 2 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 + (\sqrt{3})^2 \right] - 6 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{1}{\sqrt{3}}\right)^2 \right] \)
\( = 2 \left( \frac{1}{2} + 3 \right) - 6 \left( \frac{1}{2} - \frac{1}{3} \right) = 2 \left( \frac{7}{2} \right) - 6 \left( \frac{1}{6} \right) = 7 - 1 = 6 \)
(ii) \( 2(\cos^4 60^\circ + \sin^4 30^\circ) - (\tan^2 60^\circ + \cot^2 45^\circ) + 3\sec^2 30^\circ \)
\( = 2 \left[ \left(\frac{1}{2}\right)^4 + \left(\frac{1}{2}\right)^4 \right] - \left[ (\sqrt{3})^2 + (1)^2 \right] + 3 \left( \frac{2}{\sqrt{3}} \right)^2 \)
\( = 2 \left( \frac{1}{16} + \frac{1}{16} \right) - (3 + 1) + 3 \times \frac{4}{3} = 2 \times \frac{1}{8} - 4 + 4 = \frac{1}{4} \)

Question. Find the value of x in each of the following :
(i) \( \tan 3x = \sin 45^\circ \cos 45^\circ + \sin 30^\circ \)
(ii) \( \cos x = \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ \)
Answer: (i) \( \tan 3x = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + \frac{1}{2} \implies \tan 3x = \frac{1}{2} + \frac{1}{2} = 1 \)
\( \tan 3x = \tan 45^\circ \implies 3x = 45^\circ \implies x = 15^\circ \)
(ii) \( \cos x = \frac{1}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \times \frac{1}{2} \implies \cos x = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)
\( \cos x = \cos 30^\circ \implies x = 30^\circ \)

Question. If \( x = 30^\circ \), verify that
(i) \( \tan 2x = \frac{2\tan x}{1 - \tan^2 x} \)
(ii) \( \sin x = \sqrt{\frac{1 - \cos 2x}{2}} \)
Answer: (i) When \( x = 30^\circ, 2x = 60^\circ \).
\( \text{LHS} = \tan 60^\circ = \sqrt{3} \)
\( \text{RHS} = \frac{2\tan 30^\circ}{1 - \tan^2 30^\circ} = \frac{2(1/\sqrt{3})}{1 - (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{2/3} = \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} \). Verified.
(ii) \( \text{RHS} = \sqrt{\frac{1 - \cos 60^\circ}{2}} = \sqrt{\frac{1 - 1/2}{2}} = \sqrt{\frac{1/2}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \)
\( \text{LHS} = \sin 30^\circ = \frac{1}{2} \). Verified.

Question. Find the value of \( \theta \) in each of the following :
(i) \( 2\sin 2\theta = \sqrt{3} \)
(ii) \( 2\cos 3\theta = 1 \)
Answer: (i) \( \sin 2\theta = \frac{\sqrt{3}}{2} \implies \sin 2\theta = \sin 60^\circ \implies 2\theta = 60^\circ \implies \theta = 30^\circ \)
(ii) \( \cos 3\theta = \frac{1}{2} \implies \cos 3\theta = \cos 60^\circ \implies 3\theta = 60^\circ \implies \theta = 20^\circ \)

Question. Using the formula, \( \sin(A - B) = \sin A \cos B - \cos A \sin B \), find the value of \( \sin 15^\circ \).
Answer: Let \( A = 45^\circ, B = 30^\circ \). Then \( A - B = 15^\circ \).
\( \sin 15^\circ = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)

Question. If \( \tan(A + B) = \sqrt{3} \) and \( \tan(A - B) = \frac{1}{\sqrt{3}} \); \( 0^\circ < A + B \le 90^\circ ; A > B \), find A and B.
Answer: \( A + B = 60^\circ \) (since \( \tan 60^\circ = \sqrt{3} \))
\( A - B = 30^\circ \) (since \( \tan 30^\circ = \frac{1}{\sqrt{3}} \))
Adding both: \( 2A = 90^\circ \implies A = 45^\circ \)
Subtracting: \( 2B = 30^\circ \implies B = 15^\circ \)

TRIGONOMETRIC RATIOS OF COMPLEMENTARY ANGLES

Complementary angles are pairs of angles whose sum is \( 90^\circ \).

Formulae :

• \( \sin(90^\circ - \theta) = \cos \theta \)
• \( \cos(90^\circ - \theta) = \sin \theta \)
• \( \tan(90^\circ - \theta) = \cot \theta \)
• \( \cot(90^\circ - \theta) = \tan \theta \)
• \( \sec(90^\circ - \theta) = \csc \theta \)
• \( \csc(90^\circ - \theta) = \sec \theta \)

Question. Evaluate \( \frac{\tan 65^\circ}{\cot 25^\circ} \).
Answer: \( \frac{\tan(90^\circ - 25^\circ)}{\cot 25^\circ} = \frac{\cot 25^\circ}{\cot 25^\circ} = 1 \)

Question. Without using trigonometric tables, evaluate the following :
(i) \( \frac{\cos 37^\circ}{\sin 53^\circ} \)
(ii) \( \frac{\sin 41^\circ}{\cos 49^\circ} \)
(iii) \( \frac{\sin 30^\circ 17'}{\cos 59^\circ 43'} \)
Answer: (i) \( \frac{\cos(90^\circ - 53^\circ)}{\sin 53^\circ} = \frac{\sin 53^\circ}{\sin 53^\circ} = 1 \)
(ii) \( \frac{\sin(90^\circ - 49^\circ)}{\cos 49^\circ} = \frac{\cos 49^\circ}{\cos 49^\circ} = 1 \)
(iii) \( \frac{\sin(90^\circ - 59^\circ 43')}{\cos 59^\circ 43'} = \frac{\cos 59^\circ 43'}{\cos 59^\circ 43'} = 1 \)

Question. Without using trigonometric tables evaluate the following :
(i) \( \sin^2 25^\circ + \sin^2 65^\circ \)
(ii) \( \cos^2 13^\circ - \sin^2 77^\circ \)
Answer: (i) \( \sin^2(90^\circ - 65^\circ) + \sin^2 65^\circ = \cos^2 65^\circ + \sin^2 65^\circ = 1 \)
(ii) \( \cos^2(90^\circ - 77^\circ) - \sin^2 77^\circ = \sin^2 77^\circ - \sin^2 77^\circ = 0 \)

Question. Without using trigonometric tables, evaluate the following :
(i) \( \frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ} - 2 \)
(ii) \( \sec 50^\circ \sin 40^\circ + \cos 40^\circ \csc 50^\circ \)
Answer: (i) \( \frac{\cot(90^\circ - 36^\circ)}{\tan 36^\circ} + \frac{\tan(90^\circ - 70^\circ)}{\cot 70^\circ} - 2 = 1 + 1 - 2 = 0 \)
(ii) \( \sec(90^\circ - 40^\circ) \sin 40^\circ + \cos 40^\circ \csc(90^\circ - 40^\circ) = \csc 40^\circ \sin 40^\circ + \cos 40^\circ \sec 40^\circ = 1 + 1 = 2 \)

Question. Express each of the following in terms of trigonometric ratios of angles between \( 0^\circ \) and \( 45^\circ \);
(i) \( \csc 69^\circ + \cot 69^\circ \)
(ii) \( \sin 81^\circ + \tan 81^\circ \)
(iii) \( \sin 72^\circ + \cot 72^\circ \)
Answer: (i) \( \csc(90^\circ - 21^\circ) + \cot(90^\circ - 21^\circ) = \sec 21^\circ + \tan 21^\circ \)
(ii) \( \sin(90^\circ - 9^\circ) + \tan(90^\circ - 9^\circ) = \cos 9^\circ + \cot 9^\circ \)
(iii) \( \sin(90^\circ - 18^\circ) + \cot(90^\circ - 18^\circ) = \cos 18^\circ + \tan 18^\circ \)

Question. If \( \tan 2\theta = \cot(\theta + 6^\circ) \), find \( \theta \).
Answer: \( \cot(90^\circ - 2\theta) = \cot(\theta + 6^\circ) \implies 90^\circ - 2\theta = \theta + 6^\circ \implies 3\theta = 84^\circ \implies \theta = 28^\circ \)

Question. If A, B, C are the interior angles of a triangle ABC, prove that \( \tan \frac{B+C}{2} = \cot \frac{A}{2} \).
Answer: \( A + B + C = 180^\circ \implies \frac{B+C}{2} = 90^\circ - \frac{A}{2} \)
\( \tan \left(\frac{B+C}{2}\right) = \tan(90^\circ - A/2) = \cot(A/2) \). Proved.

Question. Prove the following identities :
(i) \( \cos^4 A - \cos^2 A = \sin^4 A - \sin^2 A \)
(ii) \( \cot^4 A - 1 = \csc^4 A - 2\csc^2 A \)
(iii) \( \sin^6 A + \cos^6 A = 1 - 3\sin^2 A \cos^2 A \).
Answer: (i) We have, \( \text{LHS} = \cos^4 A - \cos^2 A = \cos^2 A (\cos^2 A - 1) \)
\( = -\cos^2 A (1 - \cos^2 A) = -\cos^2 A \sin^2 A \)
\( = -(1 - \sin^2 A) \sin^2 A = -\sin^2 A + \sin^4 A \)
\( = \sin^4 A - \sin^2 A = \text{RHS} \)
(ii) We have, \( \text{LHS} = \cot^4 A - 1 = (\csc^2 A - 1)^2 - 1 \)
[\( \because \cot^2 A = \csc^2 A - 1 \)]
\( \therefore \cot^4 A = (\csc^2 A - 1)^2 \)
\( = \csc^4 A - 2\csc^2 A + 1 - 1 \)
\( = \csc^4 A - 2\csc^2 A = \text{RHS} \)
(iii) We have, \( \text{LHS} = \sin^6 A + \cos^6 A = (\sin^2 A)^3 + (\cos^2 A)^3 \)
\( = (\sin^2 A + \cos^2 A) \{(\sin^2 A)^2 + (\cos^2 A)^2 - \sin^2 A \cos^2 A\} \)
[\( \because a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)]
\( = \{(\sin^2 A)^2 + (\cos^2 A)^2 + 2\sin^2 A \cos^2 A - 3\sin^2 A \cos^2 A\} \)
\( = [(\sin^2 A + \cos^2 A)^2 - 3\sin^2 A \cos^2 A] \)
\( = 1 - 3\sin^2 A \cos^2 A = \text{RHS} \)

Question. If \( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \), prove that each of the side is equal to \( \pm 1 \).
Answer: We have, \( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \)
Multiplying both sides by \( (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \) we get
\( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) \)
\( = (\sec A - \tan A)^2 (\sec B - \tan B)^2 (\sec C - \tan C)^2 \)
\( \implies (\sec^2 A - \tan^2 A)(\sec^2 B - \tan^2 B)(\sec^2 C - \tan^2 C) = [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)]^2 \)
\( \implies 1 = [(\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C)]^2 \)
\( \implies (\sec A - \tan A)(\sec B - \tan B)(\sec C - \tan C) = \pm 1 \)
Similarly, multiplying both sides by \( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) \), we get
\( (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) = \pm 1 \)