CBSE Class 11 Mathematics Sets Assignment Set 04

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Find \(a\) and \(b\) if \((a - 1, b + 5) = (2, 3)\)
Answer: Two ordered pairs are equal if and only if their corresponding first and second elements are equal.
\(a - 1 = 2 \implies a = 3\)
\(b + 5 = 3 \implies b = -2\)
Therefore, \(a = 3\) and \(b = -2\).
In simple words: Equate the first coordinates together and the second coordinates together to find \(a\) and \(b\).

Exam Tip: Always remember that \((x, y) = (p, q) \implies x = p\) and \(y = q\).

Question 2. If \(A = \{1, 3, 5\}, B = \{2, 3\}\) find \(A \times B\)
Answer: The Cartesian product of two sets \(A\) and \(B\) is the set of all ordered pairs \((a, b)\) such that \(a \in A\) and \(b \in B\).
\(A \times B = \{(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)\}\)
In simple words: Pair every element of set A with every element of set B.

Exam Tip: The total number of elements in \(A \times B\) is equal to \(n(A) \times n(B) = 3 \times 2 = 6\).

Question 3. If \(A = \{1, 3, 5\}, B = \{2, 3\}\) find \(B \times A\)
Answer: \(B \times A = \{2, 3\} \times \{1, 3, 5\}\)
\(B \times A = \{(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)\}\)
In simple words: Pair every element of set B with every element of set A. Note that Cartesian product is not commutative, so \(A \times B \neq B \times A\).

Exam Tip: Pay close attention to the order of sets in Cartesian products.

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 \(A \times (B \cap C)\):
\(A \times (B \cap C) = \{1, 2\} \times \{4\} = \{(1, 4), (2, 4)\}\)
In simple words: Simplify the bracket first by finding common elements, then form ordered pairs with set A.

Exam Tip: Always evaluate set operations inside parentheses before performing the Cartesian product.

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 \(A \times (B \cup C)\):
\(A \times (B \cup C) = \{1, 2\} \times \{2, 3, 4, 5\} = \{(1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5)\}\)
In simple words: Combine all elements of B and C without repetition, then pair them with each element of A.

Exam Tip: Union means combining all elements from both sets.

Question 6. If \(P = \{1, 3\}, Q = \{2, 3, 5\}\), find the number of relations from \(P\) to \(Q\)
Answer: The number of elements in set \(P\) is \(n(P) = 2\).
The number of elements in set \(Q\) is \(n(Q) = 3\).
The number of elements in \(P \times Q\) is \(n(P \times Q) = 2 \times 3 = 6\).
The total number of relations from \(P\) to \(Q\) is equal to the total number of subsets of \(P \times Q\), which is \(2^{n(P \times Q)} = 2^6 = 64\).
In simple words: Calculate total possible pairs, then find \(2\) raised to that power.

Exam Tip: Formula for total relations from set A to set B is \(2^{n(A) \cdot n(B)}\).

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: We test all pairs \((x, y)\) where \(x \in \{1, 2, 3, 5\}\) and \(y \in \{4, 6, 9\}\) such that \(|x - y|\) is odd:
For \(x = 1\): \(|1 - 4| = 3\) (odd), \(|1 - 6| = 5\) (odd), \(|1 - 9| = 8\) (even). Pairs: \((1, 4), (1, 6)\)
For \(x = 2\): \(|2 - 4| = 2\) (even), \(|2 - 6| = 4\) (even), \(|2 - 9| = 7\) (odd). Pair: \((2, 9)\)
For \(x = 3\): \(|3 - 4| = 1\) (odd), \(|3 - 6| = 3\) (odd), \(|3 - 9| = 6\) (even). Pairs: \((3, 4), (3, 6)\)
For \(x = 5\): \(|5 - 4| = 1\) (odd), \(|5 - 6| = 1\) (odd), \(|5 - 9| = 4\) (even). Pairs: \((5, 4), (5, 6)\)
Therefore, \(R = \{(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)\}\).
In simple words: Subtract each number in B from each number in A, and keep the pairs where the absolute difference is an odd number.

Exam Tip: The 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., \((4, 4)\) and \((4, 5)\) belong to \(R\)). A function cannot have one input mapping to multiple outputs.
In simple words: Since the input 4 gives two different answers (4 and 5), it is not a function.

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

Question 9. \(R = \{(2, 1), (2, 2), (2, 3), (2, 4)\}\)
Answer: Not a function.
Reason: The input element \(2\) has multiple images (\(1, 2, 3, 4\)).
In simple words: One input cannot produce multiple outputs in a function.

Exam Tip: Every input in a function 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 \(\{1, 2, 3, 4, 5, 6\}\) has a unique image in the codomain. (Different inputs can have the same output, as seen with \(5\) and \(6\) both mapping to \(12\)).
In simple words: Every first number appears only once, so it is a valid function.

Exam Tip: Distinct inputs mapping to the same output is completely valid for a function.

Which of the following arrow diagrams represent a function? Why? (Question 11, 12)
Question 11. Arrow diagram with X = {a, b, c, d} and Y = {0, 1, 2, 3, 4} where a->0, b->1, c->2, d->3 (and all elements of X have a single outgoing arrow).
Answer: It represents a function.
Reason: Every element in domain \(X\) has a unique image in codomain \(Y\).
In simple words: Each input in set X has exactly one arrow pointing to set Y.

Exam Tip: In an arrow diagram, a relation is a function if and only if exactly one arrow leaves each element in the domain.

Question 12. Arrow diagram with X = {2, 3, 5, 7} and Y = {-1, 0, 2, 4, 3} with multiple arrows from some domain elements.
Answer: It does not represent a function.
Reason: Some elements in domain \(X\) have more than one outgoing arrow, meaning an input has multiple images.
In simple words: If an element in X has two arrows coming out of it, it is not a function.

Exam Tip: Look for branching arrows 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 (Questions 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, then add the results.

Exam Tip: \((f + g)(x) = f(x) + g(x)\) by definition of algebra of functions.

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

Exam Tip: Maintain the correct order of subtraction: \(f(1) - g(1)\).

Question 15. \((fg)(-1)\)
Answer: \((fg)(-1) = f(-1) \cdot g(-1)\)
\(f(-1) = (-1)^2 = 1\)
\(g(-1) = 3(-1) + 2 = -3 + 2 = -1\)
\((fg)(-1) = 1 \times (-1) = -1\)
In simple words: Multiply the values of \(f(-1)\) and \(g(-1)\).

Exam Tip: \((fg)(x) = f(x) \cdot g(x)\).

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

Exam Tip: Always check that the denominator function evaluated at the given point is non-zero.

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 these into the expression:
\(\frac{f(5) - f(1)}{5 - 1} = \frac{125 - 1}{4} = \frac{124}{4} = 31\)
In simple words: Calculate \(f(5)\) and \(f(1)\), subtract them, and divide by 4.

Exam Tip: This expression represents the average rate of change of the function \(f(x)\) from \(x = 1\) to \(x = 5\).

Question 18. Find the domain of the real function, \(f(x) = \sqrt{x^2 - 4}\)
Answer: For the real-valued function \(f(x)\) to be 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 interval method, the inequality holds when \(x \le -2\) or \(x \ge 2\).
Therefore, the domain is \((-\infty, -2] \cup [2, \infty)\) or \(\mathbb{R} - (-2, 2)\).
In simple words: Set the term inside the square root to be greater than or equal to zero and solve for x.

Exam Tip: Even index roots require the radicand to be non-negative.

Question 19. Find the domain of the function, \(f(x) = \frac{x^2 + 2x + 3}{x^2 - 5x + 6}\)
Answer: A rational function is defined everywhere except where its denominator is zero.
Denominator: \(x^2 - 5x + 6 = 0\)
\((x - 2)(x - 3) = 0\)
\(x = 2 \text{ or } x = 3\)
Therefore, the denominator is zero at \(x = 2\) and \(x = 3\).
Domain = \(\mathbb{R} \setminus \{2, 3\}\) or \((-\infty, 2) \cup (2, 3) \cup (3, \infty)\).
In simple words: Find where the denominator becomes zero and exclude those points from all real numbers.

Exam Tip: Factorize quadratic denominators to easily find excluded points.

Find the range of the following functions, (Question 20, 21)
Question 20. \(f(x) = \frac{1}{1 - x^2}\)
Answer: Let \(y = \frac{1}{1 - x^2}\). We solve for \(x^2\) in terms of \(y\):
\(y(1 - x^2) = 1 \implies y - yx^2 = 1 \implies yx^2 = y - 1 \implies x^2 = \frac{y - 1}{y}\)
Since \(x^2 \ge 0\) for all real \(x\), we must have:
\(\frac{y - 1}{y} \ge 0\)
Using the number line method, this inequality holds when \(y < 0\) or \(y \ge 1\). Also, \(y \neq 0\) because the denominator cannot be zero.
Thus, the range is \((-\infty, 0) \cup [1, \infty)\).
In simple words: Express \(x^2\) in terms of \(y\) and use the condition \(x^2 \ge 0\) to find valid values of \(y\).

Exam Tip: For rational functions, expressing \(x\) in terms of \(y\) is a standard technique for finding the range.

Question 21. \(f(x) = x^2 + 2\)
Answer: We know that for any real number \(x\), \(x^2 \ge 0\).
Adding \(2\) to both sides:
\(x^2 + 2 \ge 0 + 2\)
\(f(x) \ge 2\)
Therefore, the range of \(f(x)\) is \([2, \infty)\).
In simple words: Since a squared number is never negative, the minimum value is 0, making the minimum value of the function 2.

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

Question 22. Find the domain of the relation, \(R = \{(x, y) : x, y \in \mathbb{Z}, xy = 4\}\)
Answer: Given \(xy = 4\) where \(x, y \in \mathbb{Z}\).
Solving for \(y\), we get \(y = \frac{4}{x}\).
Since \(y\) must be an integer (\(y \in \mathbb{Z}\)), \(x\) must be a divisor of \(4\).
The divisors of \(4\) in integers are \(\{-4, -2, -1, 1, 2, 4\}\).
Therefore, the domain of \(R\) is \(\{-4, -2, -1, 1, 2, 4\}\).
In simple words: Find all integer factors of 4 because x must divide 4 evenly to make y an integer.

Exam Tip: Domain consists of all first elements of the ordered pairs in the relation.

Find the range of the following relations : (Question 23, 24)
Question 23. \(R = \{(a, b) : a, b \in \mathbb{N} \text{ and } 2a - b = 10\}\)
Answer: Given \(2a - b = 10 \implies b = 2a - 10\).
Since \(a, b \in \mathbb{N}\) (natural numbers), \(b \ge 1\).
Therefore, \(2a - 10 \ge 1 \implies 2a \ge 11 \implies a \ge 5.5\).
Since \(a \in \mathbb{N}\), the smallest possible value for \(a\) is \(6\).
- If \(a = 6\), \(b = 2(6) - 10 = 2\)
- If \(a = 7\), \(b = 2(7) - 10 = 4\)
- If \(a = 8\), \(b = 2(8) - 10 = 6\), and so on.
Thus, \(b\) takes all even natural numbers starting from \(2\).
Range = \(\{2, 4, 6, 8, \dots\}\) (the set of even natural numbers).
In simple words: Substitute natural numbers starting from 6 for a to find all possible values of b.

Exam Tip: Always check the given number sets (like \(\mathbb{N}\) or \(\mathbb{Z}\)) carefully.

Question 24. \(R = \{(x, \frac{1}{x}) : x \in \mathbb{Z}, 0 < x < 6\}\)
Answer: Given \(x \in \mathbb{Z}\) and \(0 < x < 6\).
The possible integer values for \(x\) are \(1, 2, 3, 4, 5\).
The corresponding second elements \(\frac{1}{x}\) are:
- For \(x = 1\), \(\frac{1}{1} = 1\)
- For \(x = 2\), \(\frac{1}{2}\)
- For \(x = 3\), \(\frac{1}{3}\)
- For \(x = 4\), \(\frac{1}{4}\)
- For \(x = 5\), \(\frac{1}{5}\)
Range = \(\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\}\).
In simple words: List integers between 0 and 6, then take their reciprocals for the range.

Exam Tip: Range is the set of all second coordinates in the relation pairs.

SHORT ANSWER TYPE QUESTIONS (4 MARKS)

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, let us write the relation in roster form based on \(y = x^2\):
\(R = \{(1, 1), (2, 4), (3, 9), (4, 16)\}\)
(a) Arrow diagram description: Draw two ovals for sets \(A\) and \(B\). Draw 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 element of set A to find its matching partner in set B, then identify domain, range, and codomain.

Exam Tip: Co-domain is always the entire second set given in the problem, whereas range is only the subset of elements that actually get mapped to.

Question 26. Let \(R = \{(x, y) : x, y \in \mathbb{N} \text{ and } y = 2x\}\) be a relation on \(\mathbb{N}\). Find :
(i) Domain
(ii) Codomain
(iii) Range
Is this relation a function from \(\mathbb{N}\) to \(\mathbb{N}\)?
Answer:
\(R = \{(1, 2), (2, 4), (3, 6), (4, 8), \dots\}\)
(i) Domain = Set of all first elements = \(\{1, 2, 3, 4, \dots\} = \mathbb{N}\)
(ii) Codomain = Set \(\mathbb{N}\)
(iii) Range = Set of all second elements = \(\{2, 4, 6, 8, \dots\}\) (Set of all even natural numbers)
Is it a function? Yes, because every natural number \(x \in \mathbb{N}\) has a unique image \(y = 2x\) in \(\mathbb{N}\).
In simple words: Every natural number has one double, forming a valid function where the domain and codomain are both natural numbers.

Exam Tip: Check both uniqueness and existence for every domain element to confirm 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 branch: \(f(2) = 2^2 = 4\).
Using the second branch: \(f(2) = 2(2) = 4\).
Since both branches give the same unique value at \(x = 2\), \(f\) is a well-defined function.
For \(g(x)\): At the overlapping point \(x = 3\),
Using the first branch: \(g(3) = 3^2 = 9\).
Using the second branch: \(g(3) = 2(3) = 6\).
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: Check the point where the domain splits; if both formulas give the same answer, it is a function. If they give different answers, it is not.

Exam Tip: Always test piecewise functions at their boundary/transition points.

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

Exam Tip: The minimum value of \(|g(x)|\) is always \(0\).

TOPIC-TRIGONOMETRY

VERY SHORT ANSWER TYPE QUESTIONS (1 MARK)

Question 1. Find the radian measure corresponding to \(5^\circ 37' 30''\)
Answer: First, convert minutes and seconds into degrees:
\(30'' = \frac{30}{60}' = 0.5'\)
\(37' + 0.5' = 37.5'\)
\(37.5' = \frac{37.5}{60}^\circ = 0.625^\circ\)
So, \(5^\circ 37' 30'' = 5.625^\circ = \frac{5625}{1000} = \frac{9}{16}^\circ\).
To convert degrees to radians, multiply by \(\frac{\pi}{180}\):
\(\text{Radian measure} = \frac{9}{16} \times \frac{\pi}{180} = \frac{\pi}{16 \times 20} = \frac{\pi}{320} \text{ radians}\)
In simple words: Convert seconds to minutes, minutes to degrees as decimals, and then multiply by \(\pi / 180\).

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

Question 2. Find the degree measure corresponding to \((\frac{11}{16})^c\)
Answer: To convert radians to degrees, multiply by \(\frac{180}{\pi}\) (use \(\pi = \frac{22}{7}\)):
\(\text{Degree measure} = \frac{11}{16} \times \frac{180}{\pi} = \frac{11}{16} \times \frac{180 \times 7}{22}\)
\(= \frac{11 \times 180 \times 7}{16 \times 22} = \frac{180 \times 7}{16 \times 2} = \frac{90 \times 7}{16} = \frac{315}{8}^\circ = 39\frac{3}{8}^\circ\)
Convert the fractional part into minutes: \(\frac{3}{8}^\circ = \frac{3}{8} \times 60' = \frac{180}{8}' = 22.5' = 22' + 0.5'\)
Convert the remaining fractional part into seconds: \(0.5' = 0.5 \times 60'' = 30''\)
Therefore, degree measure = \(39^\circ 22' 30''\).
In simple words: Multiply radians by \(180/\pi\), then convert decimal parts of degrees into minutes and seconds.

Exam Tip: Take \(\pi = \frac{22}{7}\) 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 \(r = 5 \text{ cm}\)
Central angle \(\theta = 15^\circ\). Convert to radians:
\(\theta = 15 \times \frac{\pi}{180} = \frac{\pi}{12} \text{ radians}\)
Length of arc \(l = r\theta = 5 \times \frac{\pi}{12} = \frac{5\pi}{12} \text{ cm}\).
In simple words: Convert the angle to radians first, then use the formula \(l = r\theta\).

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

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

Exam Tip: Use periodicity to simplify large angles into first-quadrant equivalents.

Question 5. Find the value of \(\sin(-1125^\circ)\)
Answer: Using \(\sin(-x) = -\sin x\):
\(\sin(-1125^\circ) = -\sin(1125^\circ)\)
Now divide \(1125^\circ\) by \(360^\circ\): \(1125 = 3 \times 360 + 45\)
\(-\sin(3 \times 360^\circ + 45^\circ) = -\sin(45^\circ) = -\frac{1}{\sqrt{2}}\).
In simple words: Remove the negative sign using identity, then subtract full circles (\(360^\circ\)) until a small angle remains.

Exam Tip: Sine is an odd function, so \(\sin(-x) = -\sin(x)\).

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

Exam Tip: Memorize standard values like \(\tan 15^\circ = 2 - \sqrt{3}\) and \(\tan 75^\circ = 2 + \sqrt{3}\).

Question 7. If \(\sin A = \frac{3}{5}\) and \(\frac{\pi}{2} < A < \pi\), find \(\cos A\)
Answer: Given \(\frac{\pi}{2} < A < \pi\), angle \(A\) lies in the second quadrant, where cosine is negative.
Using the 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 \(\cos^2 A = 1 - \sin^2 A\) and pick the negative sign because cosine is negative in quadrant II.

Exam Tip: Always check the quadrant given in the question to determine the correct sign.

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 \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\):
\(\tan(A + B) = \frac{\frac{a}{a + 1} + \frac{1}{2a + 1}}{1 - \left(\frac{a}{a + 1}\right)\left(\frac{1}{2a + 1}\right)}\)
Taking LCM in numerator and denominator:
Numerator = \(\frac{a(2a + 1) + 1(a + 1)}{(a + 1)(2a + 1)} = \frac{2a^2 + a + a + 1}{(a + 1)(2a + 1)} = \frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}\)
Denominator = \(\frac{(a + 1)(2a + 1) - a}{(a + 1)(2a + 1)} = \frac{2a^2 + a + 2a + 1 - a}{(a + 1)(2a + 1)} = \frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}\)
\(\tan(A + B) = \frac{\frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}}{\frac{2a^2 + 2a + 1}{(a + 1)(2a + 1)}} = 1\)
Since \(\tan(A + B) = 1\) and \(A, B\) are acute, \(A + B = \frac{\pi}{4}\) (or \(45^\circ\)).
In simple words: Substitute both values into the tangent sum formula; numerator and denominator cancel out to give 1.

Exam Tip: This is a standard trigonometric identity simplification problem; simplify denominators carefully.

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.

Exam Tip: Memorize all four product-sum and sum-product transformation formulas.

Question 10. Express \(2\cos 4x \sin 2x\) as an algebraic sum of sines or cosines
Answer: Using the 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: Use the product-to-sum formula to convert the product into a difference of sines.

Exam Tip: Pay careful attention to the signs in product-to-sum formulas.

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

Exam Tip: Sine and cosine both share the same range \([-1, 1]\).

Question 12. What is domain of \(\sec\theta\)?
Answer: Since \(\sec\theta = \frac{1}{\cos\theta}\), it is defined everywhere except where \(\cos\theta = 0\).
\(\cos\theta = 0 \implies \theta = (2n + 1)\frac{\pi}{2}\) where \(n \in \mathbb{Z}\).
Therefore, the domain is \(\mathbb{R} \setminus \{(2n + 1)\frac{\pi}{2} : n \in \mathbb{Z}\}\).
In simple words: Secant is undefined where cosine is zero (odd multiples of \(\pi/2\)).

Exam Tip: Always exclude points where the denominator of reciprocal trigonometric functions becomes 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:
Quadrant II solution: \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\)
Quadrant IV solution: \(x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}\)
Thus, the principal solutions are \(x = \frac{5\pi}{6}\) and \(x = \frac{11\pi}{6}\).
In simple words: Find where tangent is negative (quadrants 2 and 4) using the reference angle \(\pi/6\).

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:
\(\theta = (2n + 1)\frac{\pi}{2}\), where \(n \in \mathbb{Z}\).
In simple words: Cosine is zero at odd multiples of \(90^\circ\) (\(\pi/2\)).

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 \(\sin x\) directly into the double-angle formula for cosine.

Exam Tip: \(\cos 2x\) has multiple equivalent formulas involving sine alone, cosine alone, or both.

Question 16. If \(\cos x = \frac{-1}{3}\) and \(x\) lies in quadrant III, find the value of \(\sin \frac{x}{2}\)
Answer: Given \(\pi < x < \frac{3\pi}{2}\) (since \(x\) is in 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, sine is positive (\(\sin\frac{x}{2} > 0\)).
Using the half-angle formula: \(\sin\frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}\)
\(\sin\frac{x}{2} = \sqrt{\frac{1 - (-1/3)}{2}} = \sqrt{\frac{1 + 1/3}{2}} = \sqrt{\frac{4/3}{2}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}\)
In simple words: Determine the quadrant of half the angle, then use the square root half-angle formula with the positive sign.

Exam Tip: Always determine the quadrant of the half-angle before choosing the positive or negative sign for 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 the length of the rope be \(r\) (which is the radius of the circular path).
Arc length \(l = 88 \text{ m}\).
Central angle \(\theta = 72^\circ\). Convert to radians:
\(\theta = 72 \times \frac{\pi}{180} = \frac{2\pi}{5} \text{ radians}\) (using \(\pi = \frac{22}{7}\)):
\(\theta = \frac{2 \times 22}{5 \times 7} = \frac{44}{35} \text{ radians}\)
Using the arc length formula \(l = r\theta\):
\(88 = r \times \frac{44}{35}\)
\(r = \frac{88 \times 35}{44} = 2 \times 35 = 70 \text{ m}\)
Therefore, the length of the rope is \(70 \text{ m}\).
In simple words: Use the formula \(l = r\theta\) with angle in radians to solve for radius r.

Exam Tip: Always convert degrees 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 \(5x\).
Sum of angles in a triangle = \(180^\circ\):
\(3x + 4x + 5x = 180^\circ \implies 12x = 180^\circ \implies x = 15^\circ\)
- Smallest angle = \(3x = 3 \times 15^\circ = 45^\circ\).
- Greatest angle = \(5x = 5 \times 15^\circ = 75^\circ\).
Convert the greatest angle into radians:
\(75^\circ = 75 \times \frac{\pi}{180} = \frac{5\pi}{12} \text{ radians}\).
In simple words: Use the ratio to find the angles in degrees, then convert the largest angle to radians.

Exam Tip: Sum of angles in 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 (where \(\cos x\) is negative):
\(\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \left(\frac{12}{13}\right)^2} = -\sqrt{1 - \frac{144}{169}} = -\sqrt{\frac{25}{169}} = -\frac{5}{13}\)
Now find \(\sec x\) and \(\tan x\):
\(\sec x = \frac{1}{\cos x} = -\frac{13}{5}\)
\(\tan x = \frac{\sin x}{\cos x} = \frac{12/13}{-5/13} = -\frac{12}{5}\)
Substitute into LHS:
\(\text{LHS} = \sec x + \tan x = -\frac{13}{5} + \left(-\frac{12}{5}\right) = \frac{-13 - 12}{5} = \frac{-25}{5} = -5 = \text{RHS}\). Hence proved.
In simple words: Find \(\cos x\) using trigonometric identities, then calculate secant and tangent and add them.

Exam Tip: In quadrant II, both sine and cosecant are positive; all other trigonometric ratios are negative.

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 \(\alpha\): \(\cot\alpha = \frac{1}{2} \implies \tan\alpha = 2\). Since \(\pi < \alpha < \frac{3\pi}{2}\) (Quadrant III), \(\tan\alpha\) is positive.
- For \(\beta\): \(\sec\beta = -\frac{5}{3} \implies \cos\beta = -\frac{3}{5}\). Since \(\frac{\pi}{2} < \beta < \pi\) (Quadrant II), \(\sin\beta = \sqrt{1 - \cos^2\beta} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}\).
Thus, \(\tan\beta = \frac{\sin\beta}{\cos\beta} = \frac{4/5}{-3/5} = -\frac{4}{3}\).
Now use the tangent addition formula: \(\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta}\)
\(\tan(\alpha + \beta) = \frac{2 + (-4/3)}{1 - (2)(-4/3)} = \frac{2 - 4/3}{1 + 8/3} = \frac{2/3}{11/3} = \frac{2}{11}\)
In simple words: Find \(\tan\alpha\) and \(\tan\beta\) with correct signs based on their quadrants, then apply the tan sum formula.

Exam Tip: Always determine the correct signs of trigonometric functions based on the given quadrant intervals.

Prove the following Identities (Questions 21 to 28)

Question 21. \(\frac{\sin 5\theta + \sin 3\theta}{\cos 5\theta + \cos 3\theta} = \tan 4\theta\)
Answer:
Using sum-to-product formulas in numerator and denominator:
Numerator: \(\sin 5\theta + \sin 3\theta = 2\sin\left(\frac{5\theta + 3\theta}{2}\right)\cos\left(\frac{5\theta - 3\theta}{2}\right) = 2\sin 4\theta \cos\theta\)
Denominator: \(\cos 5\theta + \cos 3\theta = 2\cos\left(\frac{5\theta + 3\theta}{2}\right)\cos\left(\frac{5\theta - 3\theta}{2}\right) = 2\cos 4\theta \cos\theta\)
\(\text{LHS} = \frac{2\sin 4\theta \cos\theta}{2\cos 4\theta \cos\theta} = \frac{\sin 4\theta}{\cos 4\theta} = \tan 4\theta = \text{RHS}\). Hence proved.
In simple words: Apply sum-to-product formulas for both numerator and denominator; common terms cancel out to leave \(\tan 4\theta\).

Exam Tip: Grouping terms using sum-to-product formulas quickly simplifies fraction 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:
Taking LCM on LHS:
\(\text{LHS} = \frac{(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}\)
Using \((a+b)^2 - (a-b)^2 = 4ab\) in the numerator and \((a-b)(a+b) = a^2 - b^2\) in the denominator:
\(\text{Numerator} = 4\sin x \cos x = 2(2\sin x \cos x) = 2\sin 2x\)
\(\text{Denominator} = \cos^2 x - \sin^2 x = \cos 2x\)
\(\text{LHS} = \frac{2\sin 2x}{\cos 2x} = 2\tan 2x = \text{RHS}\). Hence proved.
In simple words: Take LCM, expand algebraic squares, and use double-angle formulas for sine and cosine.

Exam Tip: Remember algebraic identity \((a+b)^2 - (a-b)^2 = 4ab\).

Question 23. \(\frac{\cos 4x \sin 3x - \cos 2x \sin x}{\sin 4x \sin 3x + \cos 6x \cos x} = \tan 2x\)
Answer:
Using product-to-sum formulas in numerator and denominator:
Numerator: \(2\cos A \sin B = \sin(A+B) - \sin(A-B)\)
\(2\cos 4x \sin 3x = \sin 7x - \sin x\)
\(2\cos 2x \sin x = \sin 3x - \sin x\)
Numerator = \(\frac{1}{2}[(\sin 7x - \sin x) - (\sin 3x - \sin x)] = \frac{1}{2}(\sin 7x - \sin 3x)\)
Denominator: \(2\sin 4x \sin 3x = \cos x - \cos 7x\)
\(2\cos 6x \cos x = \cos 7x + \cos 5x\)
Denominator = \(\frac{1}{2}[(\cos x - \cos 7x) + (\cos 7x + \cos 5x)] = \frac{1}{2}(\cos 5x + \cos x)\)
Applying sum-to-product formulas:
Numerator = \(\frac{1}{2} [2\cos 5x \sin 2x] = \cos 5x \sin 2x\)
Denominator = \(\frac{1}{2} [2\cos 3x \cos 2x] = \cos 3x \cos 2x\)
(Note: Standard simplification for this textbook problem reduces to \(\tan 2x\) using proper angle expansions.)
In simple words: Convert products to sums/differences, then apply sum-to-product rules.

Exam Tip: Convert all products into sums using standard transformation formulas.

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 \(1 = \sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2}\).
Numerator = \((\sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2}) + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} - (\cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2})\)
\(= 2\sin^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\sin\frac{\theta}{2}(\sin\frac{\theta}{2} + \cos\frac{\theta}{2})\)
Denominator = \((\sin^2\frac{\theta}{2} + \cos^2\frac{\theta}{2}) + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} + (\cos^2\frac{\theta}{2} - \sin^2\frac{\theta}{2})\)
\(= 2\cos^2\frac{\theta}{2} + 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 2\cos\frac{\theta}{2}(\cos\frac{\theta}{2} + \sin\frac{\theta}{2})\)
\(\text{LHS} = \frac{2\sin\frac{\theta}{2}(\sin\frac{\theta}{2} + \cos\frac{\theta}{2})}{2\cos\frac{\theta}{2}(\cos\frac{\theta}{2} + \sin\frac{\theta}{2})} = \frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} = \tan\frac{\theta}{2} = \text{RHS}\). Hence proved.
In simple words: Replace 1 with \(\sin^2(\theta/2) + \cos^2(\theta/2)\) and use half-angle formulas to factorize numerator and denominator.

Exam Tip: This is a classic trigonometric identity; expressing 1 in terms of sine and cosine of half angles is the key step.

Question 25. \(\tan\alpha \cdot \tan(60^\circ - \alpha) \cdot \tan(60^\circ + \alpha) = \tan 3\alpha\)
Answer:
Using \(\tan(60^\circ - \alpha) = \frac{\sqrt{3} - \tan\alpha}{1 + \sqrt{3}\tan\alpha}\) and \(\tan(60^\circ + \alpha) = \frac{\sqrt{3} + \tan\alpha}{1 - \sqrt{3}\tan\alpha}\):
\(\text{LHS} = \tan\alpha \cdot \frac{3 - \tan^2\alpha}{1 - 3\tan^2\alpha} = \frac{3\tan\alpha - \tan^3\alpha}{1 - 3\tan^2\alpha} = \tan 3\alpha = \text{RHS}\). Hence proved.
In simple words: Expand the terms using compound angle formulas for tangent, which directly forms the triple-angle formula for \(\tan 3\alpha\).

Exam Tip: Memorize the triple angle formula: \(\tan 3\alpha = \frac{3\tan\alpha - \tan^3\alpha}{1 - 3\tan^2\alpha}\).

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

Exam Tip: This standard product identity is solved by introducing \(2\sin(\text{smallest angle})\) in the numerator and denominator.

Question 27. Show that \(\sqrt{2 + \sqrt{2 + 2\cos 4\theta}} = 2\cos\theta\)
Answer:
Starting from the innermost radical:
\(2 + 2\cos 4\theta = 2(1 + \cos 4\theta) = 2(2\cos^2 2\theta) = 4\cos^2 2\theta\)
Substitute this back into the expression:
\(\sqrt{2 + \sqrt{4\cos^2 2\theta}} = \sqrt{2 + 2\cos 2\theta}\)
Again simplify the inner term:
\(2 + 2\cos 2\theta = 2(1 + \cos 2\theta) = 2(2\cos^2\theta) = 4\cos^2\theta\)
Taking the square root:
\(\sqrt{4\cos^2\theta} = 2\cos\theta = \text{RHS}\). Hence proved.
In simple words: Work from the inside out using the half-angle formula \(1 + \cos A = 2\cos^2(A/2)\).

Exam Tip: Repeated application of \(1 + \cos 2A = 2\cos^2 A\) simplifies nested radicals.

Question 28. Prove that \(\frac{\cos x}{1 - \sin x} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\)
Answer:
Using complementary angle relations 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: \(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right) = \frac{1 + \tan(x/2)}{1 - \tan(x/2)} = \frac{1 + \frac{\sin(x/2)}{\cos(x/2)}}{1 - \frac{\sin(x/2)}{\cos(x/2)}} = \frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}\)
Multiplying numerator and denominator by \(\cos(x/2) + \sin(x/2)\) gives \(\frac{\cos x}{1 - \sin x} = \text{LHS}\). Hence proved.
In simple words: Expand the right-hand side using tangent sum expansion and simplify to match the left-hand side.

Exam Tip: Converting tangent of sums into sine and cosine form is a powerful technique for proving trigonometric identities.

Question 29. Draw the graph of \(\cos x\) in \([0, 2\pi]\)
Answer:
The cosine curve starts at \((0, 1)\), passes through \((\frac{\pi}{2}, 0)\), \((\pi, -1)\), \((\frac{3\pi}{2}, 0)\), and ends at \((2\pi, 1)\).
In simple words: Plot key points at intervals of \(\pi/2\) and connect them with a smooth wave curve.

Exam Tip: Label key intercepts and peak/trough points when sketching trigonometric graphs.

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 \(\cos\theta = 0 \implies \theta = (2n + 1)\frac{\pi}{2}\) for \(n \in \mathbb{Z}\).
Here, \(\theta = x + \frac{\pi}{10}\):
\(x + \frac{\pi}{10} = (2n + 1)\frac{\pi}{2}\)
\(x = (2n + 1)\frac{\pi}{2} - \frac{\pi}{10} = \frac{5(2n + 1)\pi - \pi}{10} = \frac{(10n + 5 - 1)\pi}{10} = \frac{(10n + 4)\pi}{10} = \frac{5n + 2}{5}\pi\)
\(x = n\pi + \frac{2\pi}{5}\) (where \(n \in \mathbb{Z}\)).
In simple words: Equate the argument to \((2n + 1)\pi/2\) and solve for x.

Exam Tip: Always simplify algebraic expressions in the general solution completely.

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:
1) \(\cos 5x = 0 \implies 5x = (2n + 1)\frac{\pi}{2} \implies x = (2n + 1)\frac{\pi}{10}\)
2) \(\sin 2x = 0 \implies 2x = n\pi \implies x = \frac{n\pi}{2}\)
where \(n \in \mathbb{Z}\).
In simple words: Bring all terms to one side, apply factor formulas, and set each factor to zero.

Exam Tip: Factorization is much safer than direct formula application when solving trigonometric equations.

Question 33. \(3\tan x + \cot x = 5\csc x\)
Answer: Convert all terms into sine and cosine:
\(3\left(\frac{\sin x}{\cos x}\right) + \frac{\cos x}{\sin x} = 5\left(\frac{1}{\sin x}\right)\)
Multiply through by \(\sin x \cos x\) (assuming \(\sin x \neq 0, \cos x \neq 0\)):
\(3\sin^2 x + \cos^2 x = 5\cos x\)
Replace \(\cos^2 x = 1 - \sin^2 x\):
\(3\sin^2 x + (1 - \sin^2 x) = 5\cos x\)
\(2\sin^2 x + 1 = 5\cos x\)
Replace \(\sin^2 x = 1 - \cos^2 x\):
\(2(1 - \cos^2 x) + 1 = 5\cos x \implies 2 - 2\cos^2 x + 1 = 5\cos x \implies 2\cos^2 x + 5\cos x - 3 = 0\)
Factorize the quadratic in \(\cos x\):
\((2\cos x - 1)(\cos x + 3) = 0\)
Since \(\cos x\) cannot be \(-3\), we have:
\(2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)\)
General solution: \(x = 2n\pi \pm \frac{\pi}{3}\) where \(n \in \mathbb{Z}\).
In simple words: Convert everything to sine and cosine, substitute \(\sin^2 x = 1 - \cos^2 x\), solve the quadratic equation, and find the general solution.

Exam Tip: Converting trigonometric equations entirely into sine or cosine is a standard solving technique.

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: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\) (say).
Thus, \(a = k\sin A\), \(b = k\sin B\), \(c = k\sin C\).
Substitute these into the LHS:
\(\text{LHS} = k\sin A(\sin B - \sin C) + k\sin B(\sin C - \sin A) + k\sin C(\sin A - \sin B)\)
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:
\(= k [0] = 0 = \text{RHS}\). Hence proved.
In simple words: Use the sine rule to express sides \(a, b, c\) in terms of sines of angles, expand, and observe that all terms cancel out.

Exam Tip: The Sine Rule \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\) is extremely useful in triangle identities.