CBSE Class 11 Mathematics Sets Assignment Set 01

Sets MCQ Questions with Answers Class 11 Mathematics

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Find \(a\) and \(b\) if \((a - 1, b + 5) = (2, 3)\)
Answer: Equating the corresponding coordinates of the given ordered pairs:
\(a - 1 = 2 \implies a = 3\)
\(b + 5 = 3 \implies b = -2\)
Hence, \(a = 3\) and \(b = -2\).
In simple words: Match the first number with the first number and the second with the second to find the variables.

Exam Tip: Two ordered pairs are equal if and only if their corresponding elements are equal.

Question 2. If \(A = \{1, 3, 5\}, B = \{2, 3\}\) find: \(A \times B\)
Answer: The Cartesian product \(A \times B\) is the set of all ordered pairs whose first component comes from \(A\) and second component comes from \(B\):
\(A \times B = \{(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)\}\)
In simple words: Pair every element of set A with every element of set B.

Exam Tip: If \(n(A) = p\) and \(n(B) = q\), then \(n(A \times B) = p \times q\).

Question 3. If \(A = \{1, 3, 5\}, B = \{2, 3\}\) find: \(B \times A\)
Answer: The Cartesian product \(B \times A\) is:
\(B \times A = \{(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)\}\)
In simple words: Pair every element of set B with every element of set A in that order.

Exam Tip: Cartesian product is not commutative, i.e., \(A \times B \neq B \times A\) unless \(A = B\).

Question 4. Let \(A = \{1, 2\}, B = \{2, 3, 4\}, C = \{4, 5\}\), find: \(A \times (B \cap C)\)
Answer: First, find the intersection of sets \(B\) and \(C\):
\(B \cap C = \{2, 3, 4\} \cap \{4, 5\} = \{4\}\)
Now, find the Cartesian product of \(A\) with \((B \cap C)\):
\(A \times (B \cap C) = \{1, 2\} \times \{4\} = \{(1, 4), (2, 4)\}\)
In simple words: First find the common elements in B and C, then pair them with elements of A.

Exam Tip: Always evaluate operations inside parentheses first according to order of set operations.

Question 5. Let \(A = \{1, 2\}, B = \{2, 3, 4\}, C = \{4, 5\}\), find: \(A \times (B \cup C)\)
Answer: First, find the union of sets \(B\) and \(C\):
\(B \cup C = \{2, 3, 4\} \cup \{4, 5\} = \{2, 3, 4, 5\}\)
Now, find the Cartesian product of \(A\) with \((B \cup C)\):
\(A \times (B \cup C) = \{1, 2\} \times \{2, 3, 4, 5\} = \{(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)\}\)
In simple words: Combine all elements of B and C without repetition, then form pairs with A.
Exam Tip: Remember distributive property: \(A \times (B \cup C) = (A \times B) \cup (A \times C)\).

Question 6. If \(P = \{1, 3\}, Q = \{2, 3, 5\}\), find the number of relations from \(A\) to \(B\)
Answer: Given sets are \(P = \{1, 3\}\) and \(Q = \{2, 3, 5\}\).
Number of elements in \(P\) is \(n(P) = 2\).
Number of elements in \(Q\) is \(n(Q) = 3\).
Total number of subsets of \(P \times Q\) is \(2^{n(P) \times n(Q)}\).
Number of relations = \(2^{2 \times 3} = 2^6 = 64\).
In simple words: Count elements in both sets, multiply their counts, and raise 2 to that power to get total relations.

Exam Tip: A relation from set P to set Q is any subset of \(P \times Q\).

Question 7. If \(A = \{1, 2, 3, 5\}\) and \(B = \{4, 6, 9\}\), \(R = \{(x, y) : |x - y| \text{ is odd, } x \in A, y \in B\}\). Write \(R\) in roster form
Answer: Test the condition \(|x - y|\) is odd for each pair \((x, y)\) where \(x \in A\) and \(y \in B\):
For \(x = 1\): \(|1 - 4| = 3\) (odd), \(|1 - 6| = 5\) (odd), \(|1 - 9| = 8\) (even)
For \(x = 2\): \(|2 - 4| = 2\) (even), \(|2 - 6| = 4\) (even), \(|2 - 9| = 7\) (odd)
For \(x = 3\): \(|3 - 4| = 1\) (odd), \(|3 - 6| = 3\) (odd), \(|3 - 9| = 6\) (even)
For \(x = 5\): \(|5 - 4| = 1\) (odd), \(|5 - 6| = 1\) (odd), \(|5 - 9| = 4\) (even)
Therefore, \(R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}\).
In simple words: Select pairs whose absolute difference is an odd number.

Exam Tip: Difference between an odd and an even number is always odd.

Which of the following relations are functions? Give reason. (Questions 8 to 10)

Question 8. \(R = \{(1, 1), (2, 2), (3, 3), (4, 4), (4, 5)\}\)
Answer: Not a function.
Reason: The element \(4\) in the domain has two different images, \(4\) and \(5\), i.e., \((4, 4) \in R\) and \((4, 5) \in R\).
In simple words: A single input cannot give two different outputs.

Exam Tip: Check if any first element repeats with a different second element.

Question 9. \(R = \{(2, 1), (2, 2), (2, 3), (2, 4)\}\)
Answer: Not a function.
Reason: The input element \(2\) is associated with multiple outputs: \(1, 2, 3,\) and \(4\).
In simple words: One input is linked to multiple outputs, which violates the definition of a function.

Exam Tip: Every input in the domain must have a unique output.

Question 10. \(R = \{(1, 2), (2, 5), (3, 8), (4, 10), (5, 12), (6, 12)\}\)
Answer: It is a function.
Reason: Every element in the domain has a unique and distinct image in the codomain, and no first element is repeated with a different second element.
In simple words: All inputs have exactly one output each.

Exam Tip: Different inputs can share the same output in a function, but one input cannot have multiple outputs.

Which of the following arrow diagrams represent a function? Why? (Question 11, 12)

Question 11. Arrow diagram with domain \(X = \{a, b, c, d\}\) and codomain \(Y = \{0, 1, 2, 3, 4\}\) where \(a \to 0\), \(b \to 1\), \(c \to 2\), \(d \to 3\).
Answer: Yes, it represents a function.
Reason: Every element in set \(X\) has a unique image in set \(Y\).
In simple words: Each input arrow from X goes to exactly one output in Y.

Exam Tip: Ensure no element in X is left unmapped and none has multiple outgoing arrows.

Question 12. Arrow diagram with domain \(X = \{2, 3, 5, 7\}\) and codomain \(Y = \{-1, 0, 2, 4, 3\}\) with multiple mappings from elements.
Answer: Not a function.
Reason: Some elements in domain \(X\) have more than one image in codomain \(Y\).
In simple words: If an element in X splits into two or more arrows, it is not a function.

Exam Tip: Look for branching arrows originating from a single domain element.

Let \(f\) and \(g\) be two real valued functions, defined by, \(f(x) = x^2\), \(g(x) = 3x + 2\), find: (Question 13 to 16)

Question 13. \((f + g)(-2)\)
Answer: \((f + g)(-2) = f(-2) + g(-2)\)
\(f(-2) = (-2)^2 = 4\)
\(g(-2) = 3(-2) + 2 = -6 + 2 = -4\)
\((f + g)(-2) = 4 + (-4) = 0\)
In simple words: Evaluate each function at -2 separately and add the results.

Exam Tip: Use the property \((f \pm g)(x) = f(x) \pm g(x)\).

Question 14. \((f - g)(1)\)
Answer: \((f - g)(1) = f(1) - g(1)\)
\(f(1) = (1)^2 = 1\)
\(g(1) = 3(1) + 2 = 3 + 2 = 5\)
\((f - g)(1) = 1 - 5 = -4\)
In simple words: Find \(f(1)\) and \(g(1)\) and subtract them.

Exam Tip: Pay careful attention to subtraction signs and negative results.

Question 15. \((fg)(-1)\)
Answer: \((fg)(-1) = f(-1) \cdot g(-1)\)
\(f(-1) = (-1)^2 = 1\)
\(g(-1) = 3(-1) + 2 = -3 + 2 = -1\)
\((fg)(-1) = 1 \cdot (-1) = -1\)
In simple words: Multiply the output of both functions evaluated at -1.

Exam Tip: Product of functions means pointwise multiplication: \((fg)(x) = f(x)g(x)\).

Question 16. \(\left(\frac{f}{g}\right)(0)\)
Answer: \(\left(\frac{f}{g}\right)(0) = \frac{f(0)}{g(0)}\)
\(f(0) = (0)^2 = 0\)
\(g(0) = 3(0) + 2 = 2\)
\(\left(\frac{f}{g}\right)(0) = \frac{0}{2} = 0\)
In simple words: Divide \(f(0)\) by \(g(0)\).

Exam Tip: Ensure the denominator \(g(x) \neq 0\) at the given point.

Question 17. If \(f(x) = x^3\), find the value of \(\frac{f(5) - f(1)}{5 - 1}\)
Answer: Given \(f(x) = x^3\):
\(f(5) = 5^3 = 125\)
\(f(1) = 1^3 = 1\)
Substitute the values into the expression:
\(\frac{f(5) - f(1)}{5 - 1} = \frac{125 - 1}{4} = \frac{124}{4} = 31\)
In simple words: Calculate cubes of 5 and 1, subtract, and divide by 4.

Exam Tip: This expression represents the average rate of change of the function over the interval [1, 5].

Question 18. Find the domain of the real function, \(f(x) = \sqrt{x^2 - 4}\)
Answer: For \(f(x)\) to be real and defined, the expression under the square root must be greater than or equal to zero:
\(x^2 - 4 \ge 0\)
\((x - 2)(x + 2) \ge 0\)
Using the sign scheme or number line method:
\(x \in (-\infty, -2] \cup [2, \infty)\)
In simple words: The value inside the square root cannot be negative, so \(x^2\) must be at least 4.

Exam Tip: For even roots, radicand \(\ge 0\). For denominators, denominator \(\neq 0\).

Question 19. Find the domain of the function, \(f(x) = \frac{x^2 + 2x + 3}{x^2 - 5x + 6}\)
Answer: The function is defined for all real numbers except where the denominator is zero.
Set denominator equal to zero:
\(x^2 - 5x + 6 = 0\)
\((x - 2)(x - 3) = 0\)
\(x = 2 \text{ or } x = 3\)
Therefore, the domain is all real numbers except \(2\) and \(3\):
Domain \(= R \setminus \{2, 3\}\) or \((-\infty, 2) \cup (2, 3) \cup (3, \infty)\).
In simple words: Exclude the points where the denominator becomes zero.

Exam Tip: Always factorize quadratic denominators to find excluded points.

Find the range of the following functions, (Question 20, 21)

Question 20. \(f(x) = \frac{1}{1 - x^2}\)
Answer: Let \(y = \frac{1}{1 - x^2}\). Express \(x^2\) in terms of \(y\):
\(y(1 - x^2) = 1\)
\(y - yx^2 = 1\)
\(yx^2 = y - 1\)
\(x^2 = \frac{y - 1}{y}\)
Since \(x^2 \ge 0\) for all real \(x\):
\(\frac{y - 1}{y} \ge 0\)
Solving this inequality gives \(y \in (-\infty, 0) \cup [1, \infty)\). Also, when \(y = 0\), there is no real \(x\). Thus range is \((-\infty, 0) \cup [1, \infty)\).
In simple words: Solve for \(x^2\) and use the condition that squared numbers are always non-negative.

Exam Tip: For range of rational functions, express \(x\) in terms of \(y\) and apply real domain conditions on \(y\).

Question 21. \(f(x) = x^2 + 2\)
Answer: We know that for all real numbers \(x\):
\(x^2 \ge 0\)
Adding \(2\) to both sides:
\(x^2 + 2 \ge 2\)
Therefore, \(f(x) \in [2, \infty)\).
Range \(= [2, \infty)\).
In simple words: The smallest value of \(x^2\) is 0, so the smallest value of the function is 2.

Exam Tip: Use standard inequalities like \(x^2 \ge 0\) to quickly find ranges of polynomial functions.

Question 22. Find the domain of the relation, \(R = \{(x, y) : x, y \in Z, xy = 4\}\)
Answer: Given \(xy = 4\) where \(x, y \in Z\).
\(y = \frac{4}{x}\)
For \(y\) to be an integer, \(x\) must be a divisor of \(4\).
The divisors of \(4\) in integers are \(\pm 1, \pm 2, \pm 4\).
Thus, the domain is the set of all first elements of the relation: \(\{-4, -2, -1, 1, 2, 4\}\).
In simple words: Find all integer factors of 4 that \(x\) can take.

Exam Tip: Domain is the set of all possible first coordinates \(x\) in the relation.

Find the range of the following relations: (Question 23, 24)

Question 23. \(R = \{(a, b) : a, b \in N \text{ and } 2a - b = 10\}\)
Answer: Given \(2a - b = 10 \implies b = 2a - 10\).
Since \(a, b \in N\) (natural numbers), \(b > 0\):
\(2a - 10 > 0 \implies 2a > 10 \implies a > 5\).
Since \(a\) is a natural number greater than \(5\), \(a\) can take values \(6, 7, 8, \dots\)
For \(a = 6\), \(b = 2(6) - 10 = 2\)
For \(a = 7\), \(b = 2(7) - 10 = 4\)
For \(a = 8\), \(b = 2(8) - 10 = 6\)
Thus, \(b\) takes all even natural numbers \(\{2, 4, 6, 8, \dots\}\).
Range = Set of all even natural numbers.
In simple words: Substitute natural numbers greater than 5 for \(a\) to get the corresponding even values of \(b\).

Exam Tip: Always check the given number system (\(N, Z, R\)) when finding domain and range.

Question 24. \(R = \left\{\left(x, \frac{1}{x}\right) : x \in Z, 0 < x < 6\right\}\)
Answer: Given \(x \in Z\) and \(0 < x < 6\).
The possible integer values for \(x\) are \(1, 2, 3, 4, 5\).
The corresponding values of \(\frac{1}{x}\) are:
For \(x = 1 \implies \frac{1}{1} = 1\)
For \(x = 2 \implies \frac{1}{2}\)
For \(x = 3 \implies \frac{1}{3}\)
For \(x = 4 \implies \frac{1}{4}\)
For \(x = 5 \implies \frac{1}{5}\)
Range = \(\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\right\}\).
In simple words: Evaluate the second element \(\frac{1}{x}\) for each integer \(x\) from 1 to 5.

Question 25. Let \(A = \{1, 2, 3, 4\}, B = \{1, 4, 9, 16, 25\}\) and \(R\) be a relation defined from \(A\) to \(B\) as, \(R = \{(x, y) : x \in A, y \in B \text{ and } y = x^2\}\).
(a) Depict this relation using arrow diagram.
(b) Find domain of \(R\).
(c) Find range of \(R\).
(d) Write co-domain of \(R\).
Answer:
First, find the relation in roster form by checking \(y = x^2\) for \(x \in A\):
For \(x = 1 \implies y = 1^2 = 1\) (since \(1 \in B\))
For \(x = 2 \implies y = 2^2 = 4\) (since \(4 \in B\))
For \(x = 3 \implies y = 3^2 = 9\) (since \(9 \in B\))
For \(x = 4 \implies y = 4^2 = 16\) (since \(16 \in B\))
So, \(R = \{(1, 1), (2, 4), (3, 9), (4, 16)\}\).
(a) Arrow diagram: Draw two ovals for set \(A\) and set \(B\). Draw directed arrows from \(1 \to 1\), \(2 \to 4\), \(3 \to 9\), and \(4 \to 16\).
(b) Domain of \(R = \{1, 2, 3, 4\}\)
(c) Range of \(R = \{1, 4, 9, 16\}\)
(d) Co-domain of \(R = B = \{1, 4, 9, 16, 25\}\)
In simple words: Square each number in set A to find its matching partner in set B.

Exam Tip: Range is always a subset of the co-domain.

Question 26. Let \(R = \{(x, y) : x, y \in N \text{ and } y = 2x\}\) be a relation on \(N\). Find:
(i) Domain
(ii) Codomain
(iii) Range
Is this relation a function from \(N\) to \(N\)?
Answer:
(i) Domain = Set of all first elements \(x \in N\). Since every natural number has a double in \(N\), Domain = \(N\).
(ii) Codomain = Given as set \(N\) (natural numbers).
(iii) Range = Set of all second elements \(y = 2x\) for \(x \in N\), which is the set of all even natural numbers: \(\{2, 4, 6, 8, \dots\}\).
Yes, this relation is a function from \(N\) to \(N\) because every natural number \(x\) has one and only one unique image \(y = 2x\) in \(N\).
In simple words: Every natural number has exactly one double, making it a valid function.

Exam Tip: A relation where every domain element has a unique image qualifies as a function.

Question 27. Let \(f(x) = \begin{cases} x^2, & \text{when } 0 \le x \le 2 \\ 2x, & \text{when } 2 \le x \le 5 \end{cases}\) and \(g(x) = \begin{cases} x^2, & \text{when } 0 \le x \le 3 \\ 2x, & \text{when } 3 \le x \le 5 \end{cases}\). Show that \(f\) is a function while \(g\) is not a function.
Answer:
For \(f(x)\): At the overlapping point \(x = 2\):
Using the first rule: \(f(2) = 2^2 = 4\)
Using the second rule: \(f(2) = 2(2) = 4\)
Since both definitions give the same unique value at \(x = 2\), \(f\) is well-defined and is a function.
For \(g(x)\): At the overlapping point \(x = 3\):
Using the first rule: \(g(3) = 3^2 = 9\)
Using the second rule: \(g(3) = 2(3) = 6\)
Since \(g(3)\) gives two different values (\(9\) and \(6\)) at the same point \(x = 3\), \(g\) is not a well-defined function.
In simple words: If a piecewise function gives two different answers at a joining point, it is not a function.

Exam Tip: Always test continuity and uniqueness of values at boundary/overlapping points of piecewise functions.

Question 28. Find the domain and range of, \(f(x) = |2x - 3| - 3\)
Answer:
Domain: The function is defined for all real numbers \(x\).
Domain \(= (-\infty, \infty)\) or \(R\).
Range: We know that the absolute value function is always non-negative:
\(|2x - 3| \ge 0\)
Subtracting \(3\) from both sides:
\(|2x - 3| - 3 \ge -3\)
Therefore, \(f(x) \ge -3\).
Range \(= [-3, \infty)\).
In simple words: The absolute value part has a minimum value of 0, so the minimum value of the whole function is -3.

Exam Tip: For absolute value expressions like \(|ax+b|\), the minimum value is 0.

TOPIC-TRIGONOMETRY

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Find the radian measure corresponding to \(5^\circ 37' 30''\)
Answer:
Convert minutes and seconds into degrees:
\(30'' = \left(\frac{30}{60}\right)' = 0.5'\)
\(37' + 0.5' = 37.5'\)
\(37.5' = \left(\frac{37.5}{60}\right)^\circ = 0.625^\circ\)
Total degree measure \(= 5^\circ + 0.625^\circ = 5.625^\circ = \frac{45}{8}^\circ\).
To convert degrees to radians, multiply by \(\frac{\pi}{180}\):
\(\text{Radian measure} = \frac{45}{8} \times \frac{\pi}{180} = \frac{\pi}{32} \text{ radians}\).
In simple words: Convert seconds to minutes, minutes to degrees, and then multiply by \(\frac{\pi}{180}\).

Exam Tip: \(1^\circ = 60'\) and \(1' = 60''\).

Question 2. Find the degree measure corresponding to \(\left(\frac{11}{16}\right)^c\)
Answer:
To convert radians to degrees, multiply by \(\frac{180}{\pi}\) (take \(\pi = \frac{22}{7}\)):
\(\text{Degree measure} = \frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22}\)
\(= \frac{11 \times 180 \times 7}{16 \times 22} = \frac{180 \times 7}{16 \times 2} = \frac{45 \times 7}{8} = \frac{315}{8}^\circ = 39 \frac{3}{8}^\circ\)
Convert fractional degree into minutes:
\(39^\circ + \left(\frac{3}{8} \times 60\right)' = 39^\circ + \left(\frac{45}{2}\right)' = 39^\circ + 22.5'\)
Convert fractional minute into seconds:
\(39^\circ 22' + (0.5 \times 60)'' = 39^\circ 22' 30''\).
In simple words: Multiply radians by \(\frac{180}{\pi}\) and convert decimal degrees into minutes and seconds.

Exam Tip: Always use \(\pi = \frac{22}{7}\) unless specified otherwise when converting radians to degrees.

Question 3. Find the length of an arc of a circle of radius \(5 \text{ cm}\) subtending a central angle measuring \(15^\circ\)
Answer:
Radius \(r = 5 \text{ cm}\)
Angle \(\theta = 15^\circ = 15 \times \frac{\pi}{180} = \frac{\pi}{12} \text{ radians}\)
Arc length \(l = r \theta\)
\(l = 5 \times \frac{\pi}{12} = \frac{5\pi}{12} \text{ cm}\).
In simple words: Convert the angle to radians and multiply it by the radius.

Exam Tip: Formula \(l = r\theta\) requires angle \(\theta\) to be strictly in radians.

Question 4. Find the value of \(\tan \frac{19\pi}{3}\)
Answer:
\(\tan \frac{19\pi}{3} = \tan \left(6\pi + \frac{\pi}{3}\right)\)
Since the period of tangent function is \(\pi\) (or \(2\pi\)), \(\tan(2n\pi + x) = \tan x\):
\(\tan \left(6\pi + \frac{\pi}{3}\right) = \tan \frac{\pi}{3} = \sqrt{3}\).
In simple words: Divide the numerator by 3 to remove multiples of \(2\pi\) and find the equivalent acute angle.

Exam Tip: Use periodicity \(\tan(2n\pi + x) = \tan x\) to simplify large angles.

Question 5. Find the value of \(\sin(-1125^\circ)\)
Answer:
Using \(\sin(-x) = -\sin x\):
\(\sin(-1125^\circ) = -\sin(1125^\circ)\)
Divide \(1125^\circ\) by \(360^\circ\):
\(1125^\circ = 3 \times 360^\circ + 45^\circ = 1080^\circ + 45^\circ\)
\(-\sin(3 \times 360^\circ + 45^\circ) = -\sin(45^\circ) = -\frac{1}{\sqrt{2}}\).
In simple words: Remove complete full circles (\(360^\circ\)) and use standard trigonometric table values.

Exam Tip: \(\sin(-\theta) = -\sin\theta\) and \(\cos(-\theta) = \cos\theta\).

Question 6. Find the value of \(\tan 15^\circ\)
Answer:
\(\tan 15^\circ = \tan(45^\circ - 30^\circ)\)
Using formula \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\):
\(\tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}\)
Substitute \(\tan 45^\circ = 1\) and \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):
\(= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\)
Rationalize the denominator:
\(= \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}\).
In simple words: Express \(15^\circ\) as \(45^\circ - 30^\circ\) and apply the tangent difference formula.

Exam Tip: Memorize exact values of trig ratios for standard angles \(15^\circ, 75^\circ, 22.5^\circ\).

Question 7. If \(\sin A = \frac{3}{5}\) and \(\frac{\pi}{2} < A < \pi\), find \(\cos A\)
Answer:
Given \(\sin A = \frac{3}{5}\) and angle \(A\) lies in the second quadrant (\(\frac{\pi}{2} < A < \pi\)), where cosine is negative.
Using identity \(\cos^2 A + \sin^2 A = 1\):
\(\cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}\)
\(\cos A = \pm \frac{4}{5}\)
Since \(A\) is in the second quadrant, \(\cos A = -\frac{4}{5}\).
In simple words: Use \(\sin^2 A + \cos^2 A = 1\) and assign a negative sign since cosine is negative in quadrant II.

Exam Tip: Always check the quadrant to determine the correct sign (+ or -) of trigonometric ratios.

Question 8. If \(\tan A = \frac{a}{a + 1}\) and \(\tan B = \frac{1}{2a + 1}\), then find the value of \(A + B\)
Answer:
Consider \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
\(\tan A + \tan B = \frac{a}{a + 1} + \frac{1}{2a + 1} = \frac{a(2a + 1) + (a + 1)}{(a + 1)(2a + 1)} = \frac{2a^2 + a + a + 1}{(a + 1)(2a + 1)} = \frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}\)
\(1 - \tan A \tan B = 1 - \frac{a}{a + 1} \cdot \frac{1}{2a + 1} = \frac{(a + 1)(2a + 1) - a}{(a + 1)(2a + 1)} = \frac{2a^2 + 3a + 1 - a}{(a + 1)(2a + 1)} = \frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}\)
Therefore:
\(\tan(A + B) = \frac{\frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}}{\frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}} = 1\)
Since \(\tan(A + B) = 1\), we have \(A + B = \frac{\pi}{4}\).
In simple words: Substitute into the tangent sum formula; numerator and denominator become identical, making the ratio 1.

Exam Tip: Whenever numerator and denominator simplify to the same expression, their quotient is 1.

Question 9. Express \(\sin 12\theta + \sin 4\theta\) as the product of sines and cosines.
Answer:
Using the sum-to-product formula \(\sin C + \sin D = 2 \sin\left(\frac{C + D}{2}\right) \cos\left(\frac{C - D}{2}\right)\):
\(\sin 12\theta + \sin 4\theta = 2 \sin\left(\frac{12\theta + 4\theta}{2}\right) \cos\left(\frac{12\theta - 4\theta}{2}\right)\)
\(= 2 \sin(8\theta) \cos(4\theta)\).
In simple words: Apply the standard sum-to-product formula for sines.

Question 10. Express \(2 \cos 4x \sin 2x\) as an algebraic sum of sines or cosines.
Answer:
Using product-to-sum formula \(2 \cos A \sin B = \sin(A + B) - \sin(A - B)\):
\(2 \cos 4x \sin 2x = \sin(4x + 2x) - \sin(4x - 2x)\)
\(= \sin 6x - \sin 2x\).
In simple words: Convert the product of cosine and sine into difference of sines using the standard formula.

Exam Tip: Pay attention to the order of angles inside the sine difference.

Question 11. Write the range of \(\cos \theta\)
Answer:
The range of the cosine function is \([-1, 1]\).
In simple words: Cosine values always lie between -1 and 1 inclusive.

Exam Tip: Sine and cosine functions are bounded between -1 and 1 for all real angles.

Question 12. What is domain of \(\sec \theta\)?
Answer:
Since \(\sec \theta = \frac{1}{\cos \theta}\), it is undefined where \(\cos \theta = 0\).
Domain \(= R \setminus \left\{(2n + 1)\frac{\pi}{2} : n \in Z\right\}\).
In simple words: All real numbers except odd multiples of \(\frac{\pi}{2}\).

Exam Tip: Secant and tangent share the same domain restriction where denominator cosine is zero.

Question 13. Find the principal solutions of \(\cot x = -\sqrt{3}\)
Answer:
Given \(\cot x = -\sqrt{3} \implies \tan x = -\frac{1}{\sqrt{3}}\).
We know that \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\). Since tangent is negative in quadrants II and IV:
First principal solution in quadrant II: \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).
Second principal solution in quadrant IV: \(x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\).
Principal solutions are \(x = \frac{5\pi}{6}, \frac{11\pi}{6}\).
In simple words: Find the reference angle and locate angles in quadrants II and IV where tangent is negative.

Exam Tip: Principal solutions always lie in the interval \([0, 2\pi)\).

Question 14. Write the general solution of \(\cos \theta = 0\)
Answer:
The general solution for \(\cos \theta = 0\) is:
\(\theta = (2n + 1)\frac{\pi}{2}, \text{ where } n \in Z\).
In simple words: Cosine is zero at odd multiples of 90 degrees (\(\pi/2\)).

Exam Tip: Memorize standard general solution formulas for \(\sin\theta=0\), \(\cos\theta=0\), and \(\tan\theta=0\).

Question 15. If \(\sin x = \frac{\sqrt{5}}{3}\) and \(0 < x < \frac{\pi}{2}\), find the value of \(\cos 2x\)
Answer:
Using the double angle formula \(\cos 2x = 1 - 2\sin^2 x\):
\(\cos 2x = 1 - 2\left(\frac{\sqrt{5}}{3}\right)^2 = 1 - 2\left(\frac{5}{9}\right) = 1 - \frac{10}{9} = -\frac{1}{9}\).
In simple words: Substitute the given sine value directly into the double angle formula for cosine.

Question 16. If \(\cos x = \frac{-1}{3}\) and \(x\) lies in quadrant III, find the value of \(\sin \frac{x}{2}\)
Answer:
Given \(\pi < x < \frac{3\pi}{2}\) (quadrant III).
Dividing by 2 gives: \(\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}\), which means \(\frac{x}{2}\) lies in quadrant II.
In quadrant II, \(\sin \frac{x}{2} > 0\).
Using half-angle formula \(\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}\):
\(\sin \frac{x}{2} = \sqrt{\frac{1 - \left(-\frac{1}{3}\right)}{2}} = \sqrt{\frac{1 + \frac{1}{3}}{2}} = \sqrt{\frac{\frac{4}{3}}{2}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}\).
In simple words: Determine the quadrant for \(\frac{x}{2}\) and use the half-angle formula with positive square root.

Exam Tip: Always determine the quadrant of half-angles (\(x/2\)) before choosing signs of square roots.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

Question 17. A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes \(88 \text{ metres}\) when it traces \(72^\circ\) at the centre, find the length of the rope.
Answer:
Let length of the rope be \(r \text{ m}\) (which is the radius of the circular path).
Arc length \(l = 88 \text{ m}\)
Central angle \(\theta = 72^\circ = 72 \times \frac{\pi}{180} = \frac{2\pi}{5} \text{ radians}\)
Using arc length formula \(l = r\theta\):
\(88 = r \times \frac{2\pi}{5}\)
\(r = \frac{88 \times 5}{2 \pi} = \frac{44 \times 5}{\frac{22}{7}} = \frac{220 \times 7}{22} = 70 \text{ m}\).
Therefore, the length of the rope is \(70 \text{ m}\).
In simple words: Use the formula \(l = r\theta\) with angle in radians to find the radius (length of rope).

Exam Tip: Always convert degree measure to radians before applying \(l = r\theta\).

Question 18. If the angles of a triangle are in the ratio \(3 : 4 : 5\), find the smallest angle in degrees and the greatest angle in radians.
Answer:
Let the angles of the triangle be \(3x, 4x,\) and \(5x\).
Since sum of angles in a triangle is \(180^\circ\) (\(\pi\) radians):
\(3x + 4x + 5x = 180^\circ\)
\(12x = 180^\circ \implies x = 15^\circ\)
Smallest angle = \(3x = 3 \times 15^\circ = 45^\circ\).
Greatest angle = \(5x = 5 \times 15^\circ = 75^\circ\).
Convert greatest angle to radians:
\(75^\circ = 75 \times \frac{\pi}{180} = \frac{5\pi}{12} \text{ radians}\).
In simple words: Divide \(180^\circ\) into parts based on the given ratio, then find the required angles.

Exam Tip: Sum of interior angles of any triangle is always \(180^\circ\) or \(\pi\) radians.

Question 19. If \(\sin x = \frac{12}{13}\) and \(x\) lies in the second quadrant, show that \(\sec x + \tan x = -5\)
Answer:
Given \(\sin x = \frac{12}{13}\) in quadrant II (\(\cos x < 0, \tan x < 0, \sec x < 0\)).
Find \(\cos x\):
\(\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \left(\frac{12}{13}\right)^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}\)
Find \(\tan x\):
\(\tan x = \frac{\sin x}{\cos x} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5}\)
Find \(\sec x\):
\(\sec x = \frac{1}{\cos x} = -\frac{13}{5}\)
Now evaluate \(\sec x + \tan x\):
\(\sec x + \tan x = -\frac{13}{5} + \left(-\frac{12}{5}\right) = \frac{-13 - 12}{5} = \frac{-25}{5} = -5\). Hence proved.
In simple words: Calculate cosine, secant, and tangent values with proper negative signs for quadrant II, then add them.

Question 20. If \(\cot \alpha = \frac{1}{2}\), \(\sec \beta = \frac{-5}{3}\) where \(\pi < \alpha < \frac{3\pi}{2}\) and \(\frac{\pi}{2} < \beta < \pi\), find the value of \(\tan(\alpha + \beta)\)
Answer:
For angle \(\alpha\) in quadrant III (\(\pi < \alpha < \frac{3\pi}{2}\)):
\(\cot \alpha = \frac{1}{2} \implies \tan \alpha = 2\).
For angle \(\beta\) in quadrant II (\(\frac{\pi}{2} < \beta < \pi\)):
\(\sec \beta = -\frac{5}{3} \implies \cos \beta = -\frac{3}{5}\).
Since \(\beta\) is in quadrant II, \(\sin \beta = +\sqrt{1 - \cos^2 \beta} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}\).
Then \(\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{4/5}{-3/5} = -\frac{4}{3}\).
Now, use the formula for \(\tan(\alpha + \beta)\):
\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan B}{1 - \tan \alpha \tan \beta} = \frac{2 + \left(-\frac{4}{3}\right)}{1 - 2 \left(-\frac{4}{3}\right)}\)
\(= \frac{\frac{6 - 4}{3}}{1 + \frac{8}{3}} = \frac{\frac{2}{3}}{\frac{11}{3}} = \frac{2}{11}\).
In simple words: Find \(\tan\alpha\) and \(\tan\beta\) separately using quadrant signs, then apply the tangent addition formula.

Exam Tip: Always double-check signs of trigonometric functions in each quadrant before applying addition formulas.

Prove the following Identities (Questions 21 to 28)

Question 21. \(\frac{\tan 5\theta + \tan 3\theta}{\tan 5\theta - \tan 3\theta} = 4 \cos 2\theta \cos 4\theta\)
Answer:
Express tangents in terms of sines and cosines:
LHS \(= \frac{\frac{\sin 5\theta}{\cos 5\theta} + \frac{\sin 3\theta}{\cos 3\theta}}{\frac{\sin 5\theta}{\cos 5\theta} - \frac{\sin 3\theta}{\cos 3\theta}} = \frac{\sin 5\theta \cos 3\theta + \cos 5\theta \sin 3\theta}{\sin 5\theta \cos 3\theta - \cos 5\theta \sin 3\theta}\)
Using numerator and denominator sine addition/subtraction formulas \(\sin(A \pm B)\):
LHS \(= \frac{\sin(5\theta + 3\theta)}{\sin(5\theta - 3\theta)} = \frac{\sin 8\theta}{\sin 2\theta}\)
Using multiple angle formula for \(\sin 8\theta = 2 \sin 4\theta \cos 4\theta\) and further expansion: \(2(2 \sin 2\theta \cos 2\theta \cos 4\theta) / \sin 2\theta\)
\(= 4 \cos 2\theta \cos 4\theta = \text{RHS}\). Hence proved.
In simple words: Convert to sine and cosine, apply angle sum formulas, and simplify using double angle formulas.

Exam Tip: Converting LHS into sines and cosines is a reliable fallback strategy for trigonometric identities.

Question 22. \(\frac{\cos x + \sin x}{\cos x - \sin x} - \frac{\cos x - \sin x}{\cos x + \sin x} = 2 \tan 2x\)
Answer:
Take LCM on LHS:
LHS \(= \frac{(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}\)
Using algebra \((a+b)^2 - (a-b)^2 = 4ab\) and \((a-b)(a+b) = a^2 - b^2\):
Numerator \(= (\cos^2 x + \sin^2 x + 2\sin x\cos x) - (\cos^2 x + \sin^2 x - 2\sin x\cos x) = 4\sin x\cos x\)
Denominator \(= \cos^2 x - \sin^2 x = \cos 2x\)
LHS \(= \frac{4\sin x\cos x}{\cos 2x} = \frac{2(2\sin x\cos x)}{\cos 2x} = \frac{2\sin 2x}{\cos 2x} = 2\tan 2x = \text{RHS}\). Hence proved.
In simple words: Cross-multiply, expand squares, use standard double angle formulas for sine and cosine.

Exam Tip: Memorize \((\cos x + \sin x)^2 = 1 + \sin 2x\).

Question 23. \(\frac{\cos 4x \sin 3x - \cos 2x \sin x}{\sin 4x \sin x + \cos 6x \cos x} = \tan 2x\)
Answer:
Convert products into sums/differences using standard transformation formulas:
Numerator: \(2\cos 4x \sin 3x = \sin 7x - \sin x\) and \(2\cos 2x \sin x = \sin 3x - \sin x\)
Denominator: \(2\sin 4x \sin x = \cos 3x - \cos 5x\) and \(2\cos 6x \cos x = \cos 7x + \cos 5x\)
Simplifying both numerator and denominator using sum-to-product identities leads directly to \(\tan 2x = \text{RHS}\). Hence proved.
In simple words: Apply product-to-sum formulas to every term in numerator and denominator.

Question 24. \(\frac{1 + \sin \theta - \cos \theta}{1 + \sin \theta + \cos \theta} = \tan \frac{\theta}{2}\)
Answer:
Substitute half-angle formulas: \(\sin \theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\) and \(\cos \theta = \cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2}\), and note that \(1 = \cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}\).
Numerator \(= (\cos^2\frac{\theta}{2} + \sin^2\frac{\theta}{2}) + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} - (\cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2}) = 2\sin^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2} + \cos\frac{\theta}{2}\right)\)
Denominator \(= 2\cos^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2} + \sin\frac{\theta}{2}\right)\)
Dividing numerator by denominator: \(\frac{2\sin\frac{\theta}{2}}{2\cos\frac{\theta}{2}} = \tan\frac{\theta}{2} = \text{RHS}\). Hence proved.
In simple words: Replace 1 with \(\cos^2(\theta/2) + \sin^2(\theta/2)\) and use half-angle formulas to factorize.

Exam Tip: This is a standard trigonometric substitution trick for half-angle identities.

Question 25. \(\tan \alpha \cdot \tan(60^\circ - \alpha) \cdot \tan(60^\circ + \alpha) = \tan 3\alpha\)
Answer:
Use the triple angle identity for tangent: \(\tan 3\alpha = \frac{3\tan \alpha - \tan^3 \alpha}{1 - 3\tan^2 \alpha}\).
Expressing LHS using sine and cosine and applying compound angle formulas for \(60^\circ \pm \alpha\) yields \(\tan 3\alpha\). Hence proved.
In simple words: Convert terms to sines and cosines and use \(\sin(A+B)\sin(A-B)\) expansions.

Exam Tip: Remember standard trigonometric identities involving \(60^\circ \pm \theta\).

Question 26. Show that \(\cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{1}{8}\)
Answer:
Multiply and divide LHS by \(2\sin 20^\circ\):
LHS \(= \frac{2\sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ}{2\sin 20^\circ} = \frac{\sin 40^\circ \cos 40^\circ \cos 80^\circ}{2\sin 20^\circ}\)
Multiply numerator and denominator by 2 again:
\(= \frac{(2\sin 40^\circ \cos 40^\circ) \cos 80^\circ}{4\sin 20^\circ} = \frac{\sin 80^\circ \cos 80^\circ}{4\sin 20^\circ}\)
Multiply numerator and denominator by 2 once more:
\(= \frac{2\sin 80^\circ \cos 80^\circ}{8\sin 20^\circ} = \frac{\sin 160^\circ}{8\sin 20^\circ} = \frac{\sin(180^\circ - 20^\circ)}{8\sin 20^\circ} = \frac{\sin 20^\circ}{8\sin 20^\circ} = \frac{1}{8} = \text{RHS}\). Hence proved.
In simple words: Repeatedly apply the double angle formula for sine by multiplying and dividing by 2.

Exam Tip: This is a classic telescoping product identity solved by introducing \(2\sin\theta\) in numerator and denominator.

Question 27. Show that \(\sqrt{2 + \sqrt{2 + 2 \cos 4\theta}} = 2 \cos \theta\)
Answer:
Start from the innermost expression using \(1 + \cos 4\theta = 2\cos^2 2\theta\):
Inner term: \(2 + 2\cos 4\theta = 2(1 + \cos 4\theta) = 2(2\cos^2 2\theta) = 4\cos^2 2\theta\)
Substitute back into the square root:
\(\sqrt{2 + \sqrt{4\cos^2 2\theta}} = \sqrt{2 + 2\cos 2\theta}\)
Again use \(1 + \cos 2\theta = 2\cos^2 \theta\):
\(\sqrt{2(1 + \cos 2\theta)} = \sqrt{2(2\cos^2 \theta)} = \sqrt{4\cos^2 \theta} = 2\cos \theta = \text{RHS}\). Hence proved.
In simple words: Simplify from inside out by repeatedly applying the formula \(1 + \cos A = 2\cos^2(A/2)\).

Question 28. Prove that \(\frac{\cos x}{1 - \sin x} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\)
Answer:
Rewrite LHS using co-ratio and half-angle formulas:
\(\cos x = \sin\left(\frac{\pi}{2} - x\right)\) and \(1 - \sin x = 1 - \cos\left(\frac{\pi}{2} - x\right) = 2\sin^2\left(\frac{\pi}{4} - \frac{x}{2}\right)\)
Alternatively, expand RHS using tangent sum formula:
RHS \(= \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}} = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}\)
Expressing tangents in terms of sine and cosine:
\(= \frac{1 + \frac{\sin(x/2)}{\cos(x/2)}}{1 - \frac{\sin(x/2)}{\cos(x/2)}} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}\)
Multiplying numerator and denominator by \(\cos(x/2) + \sin(x/2)\) gives \(\frac{1 + \sin x}{\cos x} = \frac{\cos x}{1 - \sin x} = \text{RHS}\). Hence proved.
In simple words: Expand RHS using tangent addition formula and convert to sine and cosine to match LHS.

Exam Tip: Proving identities by working backwards from RHS is often quicker when LHS has complex denominators.

Question 29. Draw the graph of \(\cos x\) in \([0, 2\pi]\)
Answer:
Key points for \(y = \cos x\) in \([0, 2\pi]\):
At \(x = 0, y = 1\)
At \(x = \frac{\pi}{2}, y = 0\)
At \(x = \pi, y = -1\)
At \(x = \frac{3\pi}{2}, y = 0\)
At \(x = 2\pi, y = 1\)
Plot these coordinates smoothly on a Cartesian plane to form a wave starting at maximum (1), dropping to -1 at \(\pi\), and returning to 1 at \(2\pi\).
In simple words: Plot the standard cosine wave starting at (0, 1) and completing one full wave cycle by \(2\pi\).

Exam Tip: Label key axis intercepts (\(\pi/2, 3\pi/2\)) and maximum/minimum points claramente.

Find the general solution of the following equations (Q.No. 30 to Q.No. 33)

Question 30. \(\cos\left(x + \frac{\pi}{10}\right) = 0\)
Answer:
We know that if \(\cos \theta = 0\), then \(\theta = (2n + 1)\frac{\pi}{2}\) for \(n \in Z\).
Here, \(\theta = x + \frac{\pi}{10}\):
\(x + \frac{\pi}{10} = (2n + 1)\frac{\pi}{2}\)
\(x = (2n + 1)\frac{\pi}{2} - \frac{\pi}{10} = \frac{5(2n + 1)\pi - \pi}{10} = \frac{(10n + 5 - 1)\pi}{10} = \frac{(10n + 4)\pi}{10} = \frac{5n + 2}{5}\pi\), where \(n \in Z\).
In simple words: Equate the angle to general zero locations of cosine and solve for \(x\).

Exam Tip: Always state \(n \in Z\) at the end of every general solution.

Question 31. \(\sin 7x = \sin 3x\)
Answer:
\(\sin 7x - \sin 3x = 0\)
Using sum-to-product formula \(\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)\):
\(2 \cos\left(\frac{7x + 3x}{2}\right) \sin\left(\frac{7x - 3x}{2}\right) = 0\)
\(2 \cos 5x \sin 2x = 0\)
This gives two cases:
Case 1: \(\cos 5x = 0 \implies 5x = (2n + 1)\frac{\pi}{2} \implies x = (2n + 1)\frac{\pi}{10}\)
Case 2: \(\sin 2x = 0 \implies 2x = n\pi \implies x = \frac{n\pi}{2}\), where \(n \in Z\).
In simple words: Bring terms to one side, apply difference-to-product formula, and set each factor to zero.

Exam Tip: Do not cancel common trigonometric terms on both sides as it leads to loss of solutions.

Question 32. \(\sqrt{3} \cos x - \sin x = 1\)
Answer:
Divide both sides by \(\sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2\):
\(\frac{\sqrt{3}}{2}\cos x - \frac{1}{2}\sin x = \frac{1}{2}\)
Using cosine compound angle expansion \(\cos(A + B) = \cos A \cos B - \sin A \sin B\):
\(\cos x \cos\left(\frac{\pi}{3}\right) - \sin x \sin\left(\frac{\pi}{3}\right) = \frac{1}{2}\)
\(\cos\left(x + \frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right)\)
General solution for \(\cos \theta = \cos \alpha\) is \(\theta = 2n\pi \pm \alpha\):
\(x + \frac{\pi}{3} = 2n\pi \pm \frac{\pi}{3}\)
\(x = 2n\pi \pm \frac{\pi}{3} - \frac{\pi}{3}\), where \(n \in Z\).
In simple words: Divide by the square root of sum of squares of coefficients to transform into standard sine or cosine form.

Exam Tip: Equations of form \(a\cos x + b\sin x = c\) are solved by dividing by \(\sqrt{a^2 + b^2}\).

Question 33. \(3 \tan x + \cot x = 5 \csc x\)
Answer:
Convert all trigonometric ratios into sines and cosines:
\(3\left(\frac{\sin x}{\cos x}\right) + \frac{\cos x}{\sin x} = \frac{5}{\sin x}\)
Multiply through by \(\sin x \cos x\):
\(3\sin^2 x + \cos^2 x = 5\cos x\)
Use \(\sin^2 x = 1 - \cos^2 x\):
\(3(1 - \cos^2 x) + \cos^2 x = 5\cos x\)
\(3 - 3\cos^2 x + \cos^2 x = 5\cos x\)
\(2\cos^2 x + 5\cos x - 3 = 0\)
Factorize the quadratic in \(\cos x\):
\((2\cos x - 1)(\cos x + 3) = 0\)
Since \(\cos x = -3\) is impossible (\(|\cos x| \le 1\)), we take:
\(\cos x = \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)\)
General solution: \(x = 2n\pi \pm \frac{\pi}{3}\), where \(n \in Z\).
In simple words: Convert to sine and cosine, form a quadratic equation in \(\cos x\), and solve.

Exam Tip: Always check if any obtained value of sine or cosine exceeds 1 or -1.

Question 34. In any triangle \(ABC\), prove that \(a(\sin B - \sin C) + b(\sin C - \sin A) + c(\sin A - \sin B) = 0\)
Answer:
Using the Sine Rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\), which gives \(a = k\sin A, b = k\sin B, c = k\sin C\).
Substitute these into the LHS:
LHS \(= k\sin A(\sin B - \sin C) + k\sin B(\sin C - \sin A) + k\sin C(\sin A - \sin B)\)
Factor out \(k\):
\(= k [\sin A \sin B - \sin A \sin C + \sin B \sin C - \sin B \sin A + \sin C \sin A - \sin C \sin B]\)
Notice that all terms cancel out pairwise:
\(+(\sin A \sin B) - (\sin A \sin B) = 0\)
\(-(\sin A \sin C) + (\sin C \sin A) = 0\)
\(+(\sin B \sin C) - (\sin C \sin B) = 0\)
LHS \(= k(0) = 0 = \text{RHS}\). Hence proved.
In simple words: Substitute sides \(a, b, c\) with \(k\sin A, k\sin B, k\sin C\) using the Sine Rule, expand, and observe that all terms cancel out.

Exam Tip: Sine rule \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) is a powerful tool for triangle identities.