CBSE Class 10 Mathematics Some Applications of Trigonometry Worksheet Set 08

Question 1. If \cot \theta = 15/8, evaluate \( \frac{(2 + 2\sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(2 - 2\cos\theta)} \)
Answer: We can simplify the given trigonometric expression by factoring out 2 from both the numerator and the denominator: \[ \frac{2(1 + \sin\theta)(1 - \sin\theta)}{2(1 + \cos\theta)(1 - \cos\theta)} \] The common factor of 2 cancels out. Next, we apply the algebraic identity \( (a + b)(a - b) = a^2 - b^2 \): \[ \frac{1 - \sin^2\theta}{1 - \cos^2\theta} \] Using the fundamental Pythagorean identities, where \( 1 - \sin^2\theta = \cos^2\theta \) and \( 1 - \cos^2\theta = \sin^2\theta \), we get: \[ \frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta \] Since we are given that \( \cot\theta = \frac{15}{8} \), we substitute this value into our simplified expression: \[ \cot^2\theta = \left(\frac{15}{8}\right)^2 = \frac{225}{64} \] Thus, the value of the expression is \( \frac{225}{64} \).
In simple words: Simplifying the expression shows that it is equivalent to \( \cot^2\theta \). Squaring the value of \( \cot\theta \) gives the final result.

Exam Tip: Always simplify trigonometric expressions using algebraic and Pythagorean identities before substituting any numerical values. This saves time and avoids complex calculations with fractions.

Question 2. If 7 \sin^2\theta + 3 \cos^2\theta = 4, show that \tan\theta = 1/\sqrt{3}
Answer: We can express the entire equation in terms of one trigonometric ratio using the identity \( \cos^2\theta = 1 - \sin^2\theta \). Substituting this into the given equation: \[ 7 \sin^2\theta + 3(1 - \sin^2\theta) = 4 \] \[ 7 \sin^2\theta + 3 - 3\sin^2\theta = 4 \] \[ 4 \sin^2\theta + 3 = 4 \] \[ 4 \sin^2\theta = 1 \] \[ \sin^2\theta = \frac{1}{4} \] Assuming \( \theta \) is an acute angle, we take the positive square root: \[ \sin\theta = \frac{1}{2} \] Using \( \cos^2\theta = 1 - \sin^2\theta \), we find: \[ \cos^2\theta = 1 - \frac{1}{4} = \frac{3}{4} \implies \cos\theta = \frac{\sqrt{3}}{2} \] Now, we calculate \( \tan\theta \): \[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} \] Thus, it is shown that \( \tan\theta = \frac{1}{\sqrt{3}} \).
In simple words: By replacing \( \cos^2\theta \) with \( 1 - \sin^2\theta \), we solve for \( \sin\theta \) to get \( 1/2 \). Dividing \( \sin\theta \) by \( \cos\theta \) gives the desired value of \( \tan\theta \).

Exam Tip: Expressing equations in terms of a single trigonometric ratio (either sine or cosine) makes it much easier to solve. Always remember to check if your angle is acute when taking square roots.

Question 3. Evaluate: \tan^2 60^\circ - 2 \cos^2 60^\circ - \frac{3}{4} \sin^2 45^\circ - 4 \sin^2 30^\circ
Answer: We substitute the standard values of the trigonometric ratios: \( \tan 60^\circ = \sqrt{3} \) \( \cos 60^\circ = \frac{1}{2} \) \( \sin 45^\circ = \frac{1}{\sqrt{2}} \) \( \sin 30^\circ = \frac{1}{2} \) Substituting these values into the given expression: \[ (\sqrt{3})^2 - 2\left(\frac{1}{2}\right)^2 - \frac{3}{4}\left(\frac{1}{\sqrt{2}}\right)^2 - 4\left(\frac{1}{2}\right)^2 \] \[ = 3 - 2\left(\frac{1}{4}\right) - \frac{3}{4}\left(\frac{1}{2}\right) - 4\left(\frac{1}{4}\right) \] \[ = 3 - \frac{1}{2} - \frac{3}{8} - 1 \] \[ = 2 - \frac{1}{2} - \frac{3}{8} \] Taking 8 as the common denominator: \[ = \frac{16 - 4 - 3}{8} = \frac{9}{8} \] So, the evaluated value is \( \frac{9}{8} \).
In simple words: Substitute the known values of the standard angles into the expression and solve. Simplifying the fractions step-by-step leads to the final answer.

Exam Tip: Ensure that you double-check the squared values of standard trigonometric ratios. A common mistake is forgetting to square the values when simplifying the expression.

Question 4. Evaluate: \frac{\sec^2 54^\circ - \cot^2 36^\circ}{\csc^2 57^\circ - \tan^2 33^\circ} + 2 \sin^2 38^\circ \sec^2 52^\circ
Answer: We use complementary angle formulas to rewrite the trigonometric ratios. We know that: \( \cot 36^\circ = \tan(90^\circ - 36^\circ) = \tan 54^\circ \) \( \tan 33^\circ = \cot(90^\circ - 33^\circ) = \cot 57^\circ \) \( \sec 52^\circ = \csc(90^\circ - 52^\circ) = \csc 38^\circ \) Substituting these relations into the given expression: \[ \frac{\sec^2 54^\circ - \tan^2 54^\circ}{\csc^2 57^\circ - \cot^2 57^\circ} + 2 \sin^2 38^\circ \csc^2 38^\circ \] Using the identities \( \sec^2\theta - \tan^2\theta = 1 \) and \( \csc^2\theta - \cot^2\theta = 1 \): The first fraction simplifies to: \[ \frac{1}{1} = 1 \] Since \( \csc 38^\circ = \frac{1}{\sin 38^\circ} \), the second term simplifies to: \[ 2 \sin^2 38^\circ \left(\frac{1}{\sin^2 38^\circ}\right) = 2 \] Adding the two terms together: \[ 1 + 2 = 3 \] So, the evaluated value of the expression is 3.
In simple words: Convert the complementary ratios to express them with matching angles. This lets you use standard identities to simplify the fractions and find the final answer.

Exam Tip: Whenever you see unusual angles in a problem, look for pairs that add up to 90 degrees. Converting them using complementary relations is the most efficient way to simplify the expression.

Question 5. Evaluate: \sqrt{2} \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ
Answer: We substitute the standard trigonometric ratio values: \( \tan 45^\circ = 1 \) \( \cos 30^\circ = \frac{\sqrt{3}}{2} \) \( \sin 60^\circ = \frac{\sqrt{3}}{2} \) Substituting these values: \[ \sqrt{2}(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \] Since the last two terms are identical and have opposite signs, they cancel each other out: \[ = \sqrt{2}(1) + 0 = \sqrt{2} \] So, the evaluated value is \( \sqrt{2} \).
In simple words: Plug in the standard values for each trigonometric ratio. The terms for \( \cos^2 30^\circ \) and \( \sin^2 60^\circ \) cancel each other out, leaving \( \sqrt{2} \).

Exam Tip: Recognizing that \( \cos 30^\circ \) and \( \sin 60^\circ \) are equal can save you time because their squared values will cancel out directly in subtraction.

Question 6. If \sec^2\theta (1 + \sin\theta)(1 - \sin\theta) = k, find the value of k
Answer: We use the algebraic identity \( (1 + a)(1 - a) = 1 - a^2 \) to simplify the expression: \[ (1 + \sin\theta)(1 - \sin\theta) = 1 - \sin^2\theta \] Substituting this into the given equation: \[ \sec^2\theta (1 - \sin^2\theta) = k \] By the Pythagorean identity, we know that \( 1 - \sin^2\theta = \cos^2\theta \). Therefore: \[ \sec^2\theta \cdot \cos^2\theta = k \] Since \( \sec\theta = \frac{1}{\cos\theta} \), we have \( \sec^2\theta \cdot \cos^2\theta = 1 \). \[ k = 1 \] So, the value of \( k \) is 1.
In simple words: Multiply the two terms with sine to get \( 1 - \sin^2\theta \), which is \( \cos^2\theta \). Multiplying \( \sec^2\theta \) by \( \cos^2\theta \) gives 1.

Exam Tip: Always look out for the identity \( 1 - \sin^2\theta = \cos^2\theta \). Combining this with reciprocal relations like \( \sec\theta \cdot \cos\theta = 1 \) simplifies many trigonometric problems instantly.

Question 7. Evaluate: (\sin 90^\circ + \cos 45^\circ + \cos 60^\circ)(\cos 0^\circ - \sin 45^\circ + \sin 30^\circ)
Answer: Substitute the standard values of the trigonometric ratios: \( \sin 90^\circ = 1 \) \( \cos 45^\circ = \frac{1}{\sqrt{2}} \) \( \cos 60^\circ = \frac{1}{2} \) \( \cos 0^\circ = 1 \) \( \sin 45^\circ = \frac{1}{\sqrt{2}} \) \( \sin 30^\circ = \frac{1}{2} \) Now, substitute these into the expression: \[ \left(1 + \frac{1}{\sqrt{2}} + \frac{1}{2}\right)\left(1 - \frac{1}{\sqrt{2}} + \frac{1}{2}\right) \] Grouping the rational numbers together: \[ \left[\left(1 + \frac{1}{2}\right) + \frac{1}{\sqrt{2}}\right]\left[\left(1 + \frac{1}{2}\right) - \frac{1}{\sqrt{2}}\right] \] \[ = \left(\frac{3}{2} + \frac{1}{\sqrt{2}}\right)\left(\frac{3}{2} - \frac{1}{\sqrt{2}}\right) \] Using the algebraic identity \( (a + b)(a - b) = a^2 - b^2 \): \[ = \left(\frac{3}{2}\right)^2 - \left(\frac{1}{\sqrt{2}}\right)^2 \] \[ = \frac{9}{4} - \frac{1}{2} \] Find a common denominator: \[ = \frac{9}{4} - \frac{2}{4} = \frac{7}{4} \] So, the evaluated value is \( \frac{7}{4} \).
In simple words: Group the numbers to form a pattern of \( (a+b)(a-b) \). This lets you use the difference of squares identity to easily compute the final fraction of \( 7/4 \).

Exam Tip: When simplifying terms with irrational numbers like \( \frac{1}{\sqrt{2}} \), grouping rational terms together allows you to apply the difference of squares identity, reducing potential calculation errors.

Question 8. Find the value of: \frac{2 \sin 68^\circ}{\cos 22^\circ} - \frac{2 \cot 15^\circ}{5 \tan 75^\circ} - \frac{3 \tan 45^\circ \tan 20^\circ \tan 40^\circ \tan 50^\circ \tan 70^\circ}{5}
Answer: We can solve this by converting the complementary angles: For the first term: Since \( \cos 22^\circ = \cos(90^\circ - 68^\circ) = \sin 68^\circ \): \[ \frac{2 \sin 68^\circ}{\cos 22^\circ} = \frac{2 \sin 68^\circ}{\sin 68^\circ} = 2 \] For the second term: Since \( \tan 75^\circ = \tan(90^\circ - 15^\circ) = \cot 15^\circ \): \[ \frac{2 \cot 15^\circ}{5 \tan 75^\circ} = \frac{2 \cot 15^\circ}{5 \cot 15^\circ} = \frac{2}{5} \] For the third term: Using complementary relations: \( \tan 70^\circ = \tan(90^\circ - 20^\circ) = \cot 20^\circ \) \( \tan 50^\circ = \tan(90^\circ - 40^\circ) = \cot 40^\circ \) Also, we know \( \tan 45^\circ = 1 \). Substitute these into the numerator: \[ 3 \tan 45^\circ \tan 20^\circ \tan 40^\circ \tan 50^\circ \tan 70^\circ \] \[ = 3(1) \tan 20^\circ \tan 40^\circ \cot 40^\circ \cot 20^\circ \] Since \( \tan \theta \cdot \cot \theta = 1 \): \[ = 3 \times 1 \times 1 = 3 \] Thus, the third term is \( \frac{3}{5} \). Combining all three terms: \[ 2 - \frac{2}{5} - \frac{3}{5} = 2 - \left(\frac{2 + 3}{5}\right) = 2 - \frac{5}{5} = 2 - 1 = 1 \] So, the final value is 1.
In simple words: Convert complementary angles to simplify each fraction. Multiplying paired tangents that sum to 90 degrees equals 1, allowing the whole expression to simplify to 1.

Exam Tip: Look for pairs of angles that add up to 90 degrees (like 68 and 22, or 20 and 70). Using complementary angle formulas will help you simplify these terms quickly.

Question 9. If \sin(A + B) = 1, \cos(A - B) = 1, find A and B
Answer: Given the trigonometric equations: 1) \( \sin(A + B) = 1 \) Since \( \sin 90^\circ = 1 \), we have: \[ A + B = 90^\circ \] (Equation 1) 2) \( \cos(A - B) = 1 \) Since \( \cos 0^\circ = 1 \), we have: \[ A - B = 0^\circ \] (Equation 2) Adding Equation 1 and Equation 2: \[ (A + B) + (A - B) = 90^\circ + 0^\circ \] \[ 2A = 90^\circ \implies A = 45^\circ \] Substituting \( A = 45^\circ \) into Equation 2: \[ 45^\circ - B = 0^\circ \implies B = 45^\circ \] So, the values are \( A = 45^\circ \) and \( B = 45^\circ \).
In simple words: Use standard values to convert the trigonometric equations into linear equations. Solving the equations shows that both angles A and B are equal to 45 degrees.

Exam Tip: Ensure that you relate the standard trigonometric values to the angles correctly. Write down the two linear equations clearly before solving them simultaneously.

Question 10. If \cos(40^\circ + x) = \sin 30^\circ, find the value of x
Answer: We know that \( \sin 30^\circ = \frac{1}{2} \). Substituting this value into the given equation: \[ \cos(40^\circ + x) = \frac{1}{2} \] Since \( \cos 60^\circ = \frac{1}{2} \), we can equate the angles: \[ 40^\circ + x = 60^\circ \] \[ x = 60^\circ - 40^\circ \] \[ x = 20^\circ \] So, the value of \( x \) is \( 20^\circ \).
In simple words: Substitute the value of \( \sin 30^\circ \), which is \( 1/2 \), and find which cosine angle gives \( 1/2 \). Equating the angles gives \( x = 20^\circ \).

Exam Tip: Alternatively, you can use the complementary relationship \( \sin\theta = \cos(90^\circ - \theta) \) to write \( \sin 30^\circ = \cos 60^\circ \), and directly equate the angles.

Question 11. Sin 4A = cos (A - 20^\circ), where 4A is an acute angle, find the value of A
Answer: We use the complementary angle identity \( \sin\theta = \cos(90^\circ - \theta) \) to rewrite the left-hand side: \[ \sin 4A = \cos(90^\circ - 4A) \] Now, substitute this into the given equation: \[ \cos(90^\circ - 4A) = \cos(A - 20^\circ) \] Since both are acute angles, we can equate the terms: \[ 90^\circ - 4A = A - 20^\circ \] \[ 90^\circ + 20^\circ = A + 4A \] \[ 110^\circ = 5A \] \[ A = \frac{110^\circ}{5} = 22^\circ \] So, the value of \( A \) is \( 22^\circ \).
In simple words: Convert sine into cosine using complementary angles. Equating the two cosine angles allows us to solve for A, giving 22 degrees.

Exam Tip: When dealing with equations involving different trigonometric functions like sine and cosine, always use complementary angle formulas to convert them into the same function.

Question 12. Find the acute angles A and B, A > B, if \sin(A + 2B) = \sqrt{3}/2 and \cos(A + 4B) = 0
Answer: Given the trigonometric equations: 1) \( \sin(A + 2B) = \frac{\sqrt{3}}{2} \) Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), we have: \[ A + 2B = 60^\circ \] (Equation 1) 2) \( \cos(A + 4B) = 0 \) Since \( \cos 90^\circ = 0 \), we have: \[ A + 4B = 90^\circ \] (Equation 2) Subtracting Equation 1 from Equation 2: \[ (A + 4B) - (A + 2B) = 90^\circ - 60^\circ \] \[ 2B = 30^\circ \implies B = 15^\circ \] Now, substitute \( B = 15^\circ \) into Equation 1: \[ A + 2(15^\circ) = 60^\circ \] \[ A + 30^\circ = 60^\circ \implies A = 30^\circ \] Since \( A = 30^\circ \) and \( B = 15^\circ \), the condition \( A > B \) is satisfied. So, the acute angles are \( A = 30^\circ \) and \( B = 15^\circ \).
In simple words: Translate the trigonometric expressions into two linear equations based on standard values. Solve these equations to find that angle A is 30 degrees and B is 15 degrees.

Exam Tip: Double-check your final values against any given constraints (like \( A > B \) and that both are acute angles) to ensure your solution is correct.

Question 13. Evaluate: sec (90 - \theta)\csc\theta - \tan(90 - \theta)\cot\theta + \frac{\cos^2 35^\circ + \cos^2 55^\circ}{\tan 5^\circ \tan 15^\circ \tan 45^\circ \tan 75^\circ \tan 85^\circ}
Answer: We simplify the expression by parts: First term: \( \sec(90^\circ - \theta) \csc\theta - \tan(90^\circ - \theta) \cot\theta \) Using complementary identities: \( \sec(90^\circ - \theta) = \csc\theta \) \( \tan(90^\circ - \theta) = \cot\theta \) Substitute these in: \[ = \csc\theta \cdot \csc\theta - \cot\theta \cdot \cot\theta \] \[ = \csc^2\theta - \cot^2\theta \] By standard identity, \( \csc^2\theta - \cot^2\theta = 1 \). Second term: \( \frac{\cos^2 35^\circ + \cos^2 55^\circ}{\tan 5^\circ \tan 15^\circ \tan 45^\circ \tan 75^\circ \tan 85^\circ} \) For the numerator: Since \( \cos 55^\circ = \cos(90^\circ - 35^\circ) = \sin 35^\circ \): \[ \cos^2 35^\circ + \cos^2 55^\circ = \cos^2 35^\circ + \sin^2 35^\circ = 1 \] For the denominator: Pair the complementary angles: \( \tan 85^\circ = \tan(90^\circ - 5^\circ) = \cot 5^\circ \) \( \tan 75^\circ = \tan(90^\circ - 15^\circ) = \cot 15^\circ \) Also, \( \tan 45^\circ = 1 \). The denominator becomes: \[ \tan 5^\circ \tan 15^\circ (1) \cot 15^\circ \cot 5^\circ \] \[ = (\tan 5^\circ \cot 5^\circ)(\tan 15^\circ \cot 15^\circ) = 1 \times 1 = 1 \] Thus, the second term is \( \frac{1}{1} = 1 \). Adding the two terms: \[ 1 + 1 = 2 \] So, the evaluated value of the expression is 2.
In simple words: Simplify the first part to 1 using complementary identities and standard formulas. Pair the complementary terms in the second part to make the fraction equal to 1, giving a total of 2.

Exam Tip: Whenever you see angles like \( 35^\circ \) and \( 55^\circ \), check if they are complementary. Converting one of them using complementary angle rules simplifies the expressions easily.

Question 14. If sinA - cosB = 0, prove that A + B = 90^\circ
Answer: Given the equation: \[ \sin A - \cos B = 0 \] \[ \sin A = \cos B \] We know from complementary angle relationships that: \[ \cos B = \sin(90^\circ - B) \] Substituting this into the equation: \[ \sin A = \sin(90^\circ - B) \] Comparing the angles on both sides: \[ A = 90^\circ - B \] \[ A + B = 90^\circ \] Hence proved.
In simple words: Move \( \cos B \) to the other side of the equation and convert it to \( \sin(90^\circ - B) \). Equating the angles shows that their sum is 90 degrees.

Exam Tip: This is a standard proof. Remember that \( \sin\theta = \cos(90^\circ - \theta) \) is the key tool used to relate different trigonometric ratios in such equations.

Question 15. If \frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{5}{3}, evaluate \frac{7\tan\theta + 2}{2\tan\theta + 7}
Answer: Given the equation: \[ \frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{5}{3} \] Divide the numerator and denominator of the left-hand side by \( \cos\theta \): \[ \frac{\frac{\sin\theta}{\cos\theta} + 1}{\frac{\sin\theta}{\cos\theta} - 1} = \frac{5}{3} \] \[ \frac{\tan\theta + 1}{\tan\theta - 1} = \frac{5}{3} \] Cross-multiplying to solve for \( \tan\theta \): \[ 3(\tan\theta + 1) = 5(\tan\theta - 1) \] \[ 3\tan\theta + 3 = 5\tan\theta - 5 \] \[ 3 + 5 = 5\tan\theta - 3\tan\theta \] \[ 8 = 2\tan\theta \implies \tan\theta = 4 \] Now, substitute \( \tan\theta = 4 \) into the expression we need to evaluate: \[ \frac{7\tan\theta + 2}{2\tan\theta + 7} = \frac{7(4) + 2}{2(4) + 7} \] \[ = \frac{28 + 2}{8 + 7} = \frac{30}{15} = 2 \] So, the evaluated value is 2.
In simple words: Divide the terms in the first fraction by \( \cos\theta \) to turn them into \( \tan\theta \). Solving for \( \tan\theta \) gives 4, which we substitute into the target expression to get 2.

Exam Tip: Dividing numerator and denominator by \( \cos\theta \) is a clever technique that directly converts sine and cosine expressions into tangent, making calculations much simpler.

Question 16. What is the maximum value of 1/sec\theta
Answer: The reciprocal of the secant function is the cosine function: \[ \frac{1}{\sec\theta} = \cos\theta \] For any real angle \( \theta \), the value of \( \cos\theta \) is bounded between \( -1 \) and \( 1 \): \[ -1 \le \cos\theta \le 1 \] Therefore, the maximum possible value of \( \cos\theta \), and consequently of \( \frac{1}{\sec\theta} \), is 1.
In simple words: Since \( 1/\sec\theta \) is equal to \( \cos\theta \), and the highest value cosine can ever reach is 1, the maximum value is 1.

Exam Tip: Always rewrite reciprocal trigonometric functions as their primary counterparts (like cosine for secant) to easily analyze their limits and values.

Question 17. If A, B and C are interior angles of triangle ABC, show that \cos \left\{\frac{B+C}{2}\right\} = \sin \frac{A}{2}
Answer: In any triangle ABC, the sum of the interior angles is \( 180^\circ \): \[ A + B + C = 180^\circ \] We can express \( B + C \) in terms of \( A \): \[ B + C = 180^\circ - A \] Dividing both sides by 2: \[ \frac{B + C}{2} = \frac{180^\circ - A}{2} \] \[ \frac{B + C}{2} = 90^\circ - \frac{A}{2} \] Now, take the cosine of both sides: \[ \cos\left(\frac{B + C}{2}\right) = \cos\left(90^\circ - \frac{A}{2}\right) \] Using the complementary angle identity \( \cos(90^\circ - \theta) = \sin\theta \): \[ \cos\left(\frac{B + C}{2}\right) = \sin\left(\frac{A}{2}\right) \] Hence proved.
In simple words: Since the sum of angles in a triangle is 180 degrees, we can write \( (B+C)/2 \) as \( 90 - A/2 \). Taking cosine of this value gives \( \sin(A/2) \).

Exam Tip: In triangle-related proofs, always start with the angle sum property \( A + B + C = 180^\circ \) and rearrange it to match the terms inside the trigonometric functions.

Question 18. If x = a sin\theta, y = b tan\theta. Prove that \frac{a^2}{x^2} - \frac{b^2}{y^2} = 1
Answer: We are given: \( x = a \sin\theta \implies \frac{x}{a} = \sin\theta \) Taking the reciprocal: \[ \frac{a}{x} = \frac{1}{\sin\theta} = \csc\theta \] Squaring both sides: \[ \frac{a^2}{x^2} = \csc^2\theta \] (Equation 1) We are also given: \( y = b \tan\theta \implies \frac{y}{b} = \tan\theta \) Taking the reciprocal: \[ \frac{b}{y} = \frac{1}{\tan\theta} = \cot\theta \] Squaring both sides: \[ \frac{b^2}{y^2} = \cot^2\theta \] (Equation 2) Now, subtract Equation 2 from Equation 1: \[ \frac{a^2}{x^2} - \frac{b^2}{y^2} = \csc^2\theta - \cot^2\theta \] By the standard Pythagorean identity, we know that \( \csc^2\theta - \cot^2\theta = 1 \). Therefore: \[ \frac{a^2}{x^2} - \frac{b^2}{y^2} = 1 \] Hence proved.
In simple words: Express \( a/x \) as \( \csc\theta \) and \( b/y \) as \( \cot\theta \). Subtracting their squares gives \( \csc^2\theta - \cot^2\theta \), which always equals 1.

Exam Tip: Identify reciprocal relationships early in proof questions. Converting \( \sin\theta \) to \( \csc\theta \) and \( \tan\theta \) to \( \cot\theta \) allows you to use standard Pythagorean identities directly.

Question 19. Prove that: \frac{1}{1 + \sin\theta} + \frac{1}{1 - \sin\theta} = 2 \sec^2 \theta
Answer: We start with the Left-Hand Side (LHS) of the expression: \[ \text{LHS} = \frac{1}{1 + \sin\theta} + \frac{1}{1 - \sin\theta} \] Taking a common denominator: \[ = \frac{(1 - \sin\theta) + (1 + \sin\theta)}{(1 + \sin\theta)(1 - \sin\theta)} \] In the numerator, \( -\sin\theta \) and \( +\sin\theta \) cancel out: \[ = \frac{2}{1 - \sin^2\theta} \] Using the fundamental identity \( 1 - \sin^2\theta = \cos^2\theta \): \[ = \frac{2}{\cos^2\theta} \] Since \( \frac{1}{\cos^2\theta} = \sec^2\theta \): \[ = 2\sec^2\theta \] This is equal to the Right-Hand Side (RHS). Hence proved.
In simple words: Combine the two fractions by finding a common denominator. This gives \( 2/(1 - \sin^2\theta) \), which simplifies directly to \( 2\sec^2\theta \).

Exam Tip: When adding fractions in trigonometry, cross-multiplying to find a common denominator often leads to a difference of squares in the denominator, which simplifies using identities.

Question 20. Prove that: \frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} = 2\csc\theta
Answer: We start with the Left-Hand Side (LHS): \[ \text{LHS} = \frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} \] Find a common denominator and combine the fractions: \[ = \frac{\sin^2\theta + (1 + \cos\theta)^2}{(1 + \cos\theta)\sin\theta} \] Expand the numerator using the identity \( (a+b)^2 = a^2 + 2ab + b^2 \): \[ = \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{(1 + \cos\theta)\sin\theta} \] Using the identity \( \sin^2\theta + \cos^2\theta = 1 \), we combine those terms in the numerator: \[ = \frac{1 + 1 + 2\cos\theta}{(1 + \cos\theta)\sin\theta} \] \[ = \frac{2 + 2\cos\theta}{(1 + \cos\theta)\sin\theta} \] Factor out 2 in the numerator: \[ = \frac{2(1 + \cos\theta)}{(1 + \cos\theta)\sin\theta} \] Since \( 1 + \cos\theta \) is present in both the numerator and denominator, they cancel out: \[ = \frac{2}{\sin\theta} \] Since \( \frac{1}{\sin\theta} = \csc\theta \): \[ = 2\csc\theta \] This is equal to the Right-Hand Side (RHS). Hence proved.
In simple words: Add the fractions together and use the identity \( \sin^2\theta + \cos^2\theta = 1 \) to simplify the numerator. Factoring out 2 allows the common term to cancel, leaving \( 2/\sin\theta \).

Exam Tip: When expanding brackets like \( (1 + \cos\theta)^2 \), always group \( \sin^2\theta \) and \( \cos^2\theta \) together immediately to simplify the expression using Pythagorean identities.

Question 21. If \tan \theta + \sin \theta = m and \tan \theta - \sin \theta = n, show that (m^2 - n^2) = 4\sqrt{mn}
Answer: We evaluate the Left-Hand Side (LHS) first: \[ \text{LHS} = m^2 - n^2 = (m - n)(m + n) \] Substitute the expressions for \( m \) and \( n \): \[ m + n = (\tan\theta + \sin\theta) + (\tan\theta - \sin\theta) = 2\tan\theta \] \[ m - n = (\tan\theta + \sin\theta) - (\tan\theta - \sin\theta) = 2\sin\theta \] Multiplying these together: \[ m^2 - n^2 = (2\tan\theta)(2\sin\theta) = 4\tan\theta\sin\theta \] (Equation 1) Now, we evaluate the Right-Hand Side (RHS): \[ \text{RHS} = 4\sqrt{mn} \] Substitute \( m \) and \( n \) inside the square root: \[ mn = (\tan\theta + \sin\theta)(\tan\theta - \sin\theta) \] \[ = \tan^2\theta - \sin^2\theta \] \[ = \frac{\sin^2\theta}{\cos^2\theta} - \sin^2\theta \] Factor out \( \sin^2\theta \): \[ = \sin^2\theta \left(\frac{1}{\cos^2\theta} - 1\right) \] \[ = \sin^2\theta (\sec^2\theta - 1) \] Using the identity \( \sec^2\theta - 1 = \tan^2\theta \): \[ = \sin^2\theta \tan^2\theta \] Now substitute this back into the RHS expression: \[ 4\sqrt{mn} = 4\sqrt{\sin^2\theta \tan^2\theta} = 4\sin\theta\tan\theta \] (Equation 2) Comparing Equation 1 and Equation 2, we see that: \[ \text{LHS} = \text{RHS} \] Hence proved.
In simple words: Use the difference of squares to find \( m^2 - n^2 = 4\tan\theta\sin\theta \). Then multiply \( m \) and \( n \) inside the square root and simplify to get the same result.

Exam Tip: Converting \( \tan^2\theta - \sin^2\theta \) to \( \sin^2\theta\tan^2\theta \) inside the square root is the crucial step in this problem. Practice factoring out sine squared to make this simplification easy.

Question 22. 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: We need to eliminate angle B from the equations. From the first equation: \[ \tan A = n \tan B \implies \tan B = \frac{\tan A}{n} \] Taking the reciprocal: \[ \cot B = \frac{n}{\tan A} \] (Equation 1) From the second equation: \[ \sin A = m \sin B \implies \sin B = \frac{\sin A}{m} \] Taking the reciprocal: \[ \csc B = \frac{m}{\sin A} \] (Equation 2) We know the trigonometric identity: \[ \csc^2 B - \cot^2 B = 1 \] Substitute Equation 1 and Equation 2 into the identity: \[ \left(\frac{m}{\sin A}\right)^2 - \left(\frac{n}{\tan A}\right)^2 = 1 \] \[ \frac{m^2}{\sin^2 A} - \frac{n^2}{\tan^2 A} = 1 \] Since \( \tan^2 A = \frac{\sin^2 A}{\cos^2 A} \): \[ \frac{m^2}{\sin^2 A} - \frac{n^2 \cos^2 A}{\sin^2 A} = 1 \] Combining the fractions: \[ \frac{m^2 - n^2 \cos^2 A}{\sin^2 A} = 1 \] \[ m^2 - n^2 \cos^2 A = \sin^2 A \] Using the Pythagorean identity \( \sin^2 A = 1 - \cos^2 A \): \[ m^2 - n^2 \cos^2 A = 1 - \cos^2 A \] Rearranging terms to solve for \( \cos^2 A \): \[ m^2 - 1 = n^2 \cos^2 A - \cos^2 A \] \[ m^2 - 1 = (n^2 - 1)\cos^2 A \] \[ \cos^2 A = \frac{m^2 - 1}{n^2 - 1} \] Hence proved.
In simple words: Find \( \csc B \) and \( \cot B \) from the given equations, and substitute them into \( \csc^2 B - \cot^2 B = 1 \). Simplify the terms to solve for \( \cos^2 A \).

Exam Tip: Eliminating an angle (like B) is done by finding two expressions that fit a known trigonometric identity (such as \( \csc^2 B - \cot^2 B = 1 \)).

Question 23. 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: We are given two equations: 1) \( x \sin^3\theta + y \cos^3\theta = \sin\theta \cos\theta \) 2) \( x \sin\theta = y \cos\theta \) We can rewrite the first equation as: \[ (x \sin\theta) \sin^2\theta + (y \cos\theta) \cos^2\theta = \sin\theta \cos\theta \] Using the second equation, we substitute \( x \sin\theta \) in place of \( y \cos\theta \): \[ (x \sin\theta) \sin^2\theta + (x \sin\theta) \cos^2\theta = \sin\theta \cos\theta \] Factor out the common term \( x \sin\theta \): \[ x \sin\theta (\sin^2\theta + \cos^2\theta) = \sin\theta \cos\theta \] Since \( \sin^2\theta + \cos^2\theta = 1 \): \[ x \sin\theta (1) = \sin\theta \cos\theta \] \[ x \sin\theta = \sin\theta \cos\theta \] Dividing both sides by \( \sin\theta \) (assuming \( \sin\theta \neq 0 \)): \[ x = \cos\theta \] Now, substitute \( x = \cos\theta \) back into the second equation \( x \sin\theta = y \cos\theta \): \[ \cos\theta \sin\theta = y \cos\theta \] Dividing both sides by \( \cos\theta \) (assuming \( \cos\theta \neq 0 \)): \[ y = \sin\theta \] Now, substitute the values of \( x \) and \( y \) into the expression to be proved: \[ x^2 + y^2 = (\cos\theta)^2 + (\sin\theta)^2 \] \[ x^2 + y^2 = \cos^2\theta + \sin^2\theta = 1 \] Hence proved.
In simple words: Substitute \( x \sin\theta \) for \( y \cos\theta \) in the first equation to simplify it. This reveals that \( x = \cos\theta \) and \( y = \sin\theta \), which means \( x^2 + y^2 = 1 \).

Exam Tip: Splitting cubed terms like \( \sin^3\theta \) into \( \sin\theta \cdot \sin^2\theta \) is a useful trick that lets you substitute and simplify terms using other given equations.

Question 24. If \sin\theta + \cos\theta = \sqrt{3} then prove that \tan\theta + \cot\theta = 1.
Answer: We are given: \[ \sin\theta + \cos\theta = \sqrt{3} \] Squaring both sides of the equation: \[ (\sin\theta + \cos\theta)^2 = (\sqrt{3})^2 \] \[ \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 3 \] Since \( \sin^2\theta + \cos^2\theta = 1 \): \[ 1 + 2\sin\theta\cos\theta = 3 \] \[ 2\sin\theta\cos\theta = 2 \] \[ \sin\theta\cos\theta = 1 \] Now, let us evaluate the Left-Hand Side (LHS) of the expression we want to prove: \[ \text{LHS} = \tan\theta + \cot\theta \] Expressing tangent and cotangent in terms of sine and cosine: \[ = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} \] Taking a common denominator: \[ = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \] Using the identity \( \sin^2\theta + \cos^2\theta = 1 \): \[ = \frac{1}{\sin\theta\cos\theta} \] Substitute \( \sin\theta\cos\theta = 1 \): \[ = \frac{1}{1} = 1 \] This is equal to the RHS. Hence proved.
In simple words: Square both sides of the first equation to find that \( \sin\theta\cos\theta = 1 \). Convert the tangent and cotangent terms to sine and cosine to show their sum is also 1.

Exam Tip: Whenever you are given a sum of sine and cosine, squaring both sides is almost always the first step because it immediately produces the term \( 2\sin\theta\cos\theta \).

Question 25. 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: Let the required expression be \( x \): \[ a \sin\theta + b \cos\theta = x \] We are given: \[ a \cos\theta - b \sin\theta = c \] Square both equations: \[ (a \cos\theta - b \sin\theta)^2 = c^2 \] \[ a^2 \cos^2\theta + b^2 \sin^2\theta - 2ab\sin\theta\cos\theta = c^2 \] (Equation 1) \[ (a \sin\theta + b \cos\theta)^2 = x^2 \] \[ a^2 \sin^2\theta + b^2 \cos^2\theta + 2ab\sin\theta\cos\theta = x^2 \] (Equation 2) Adding Equation 1 and Equation 2: \[ (a^2 \cos^2\theta + a^2 \sin^2\theta) + (b^2 \sin^2\theta + b^2 \cos^2\theta) = c^2 + x^2 \] Factor out \( a^2 \) and \( b^2 \): \[ a^2(\cos^2\theta + \sin^2\theta) + b^2(\sin^2\theta + \cos^2\theta) = c^2 + x^2 \] Since \( \sin^2\theta + \cos^2\theta = 1 \): \[ a^2(1) + b^2(1) = c^2 + x^2 \] \[ a^2 + b^2 = c^2 + x^2 \] Now, solve for \( x^2 \): \[ x^2 = a^2 + b^2 - c^2 \] Taking the square root on both sides: \[ x = \pm\sqrt{a^2 + b^2 - c^2} \] Substitute \( x = a \sin\theta + b \cos\theta \): \[ a \sin\theta + b \cos\theta = \pm\sqrt{a^2 + b^2 - c^2} \] Hence proved.
In simple words: Square both equations and add them together. This cancels out the middle term and simplifies the rest to \( a^2 + b^2 = c^2 + x^2 \), which we can solve for our unknown.

Exam Tip: Adding the squares of two expressions with swapped sine/cosine coefficients (like \( a\cos\theta - b\sin\theta \) and \( a\sin\theta + b\cos\theta \)) is a standard technique that eliminates the \( 2ab\sin\theta\cos\theta \) cross-term.

Question 26. If \sin\theta + \cos\theta = a, \tan\theta + \cot\theta = b, show that \frac{a^2 - 1}{2} = \frac{1}{b}
Answer: We are given: \[ \sin\theta + \cos\theta = a \] Squaring both sides: \[ (\sin\theta + \cos\theta)^2 = a^2 \] \[ \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = a^2 \] Since \( \sin^2\theta + \cos^2\theta = 1 \): \[ 1 + 2\sin\theta\cos\theta = a^2 \] \[ 2\sin\theta\cos\theta = a^2 - 1 \] \[ \sin\theta\cos\theta = \frac{a^2 - 1}{2} \] (Equation 1) We are also given: \[ \tan\theta + \cot\theta = b \] Convert to sine and cosine: \[ \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = b \] \[ \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = b \] \[ \frac{1}{\sin\theta\cos\theta} = b \] Taking the reciprocal: \[ \sin\theta\cos\theta = \frac{1}{b} \] (Equation 2) Equating Equation 1 and Equation 2: \[ \frac{a^2 - 1}{2} = \frac{1}{b} \] Hence proved.
In simple words: Square the first equation to express \( \sin\theta\cos\theta \) as \( (a^2 - 1)/2 \). Express the second equation as \( \sin\theta\cos\theta = 1/b \). Equating both expressions proves the result.

Exam Tip: Look for common terms (like \( \sin\theta\cos\theta \)) in both given equations to link them together and complete the proof easily.

Question 27. If x = a \cos^3\theta, y = b \sin^3\theta, prove that \( \left(\frac{x}{a}\right)^{2/3} + \left(\frac{y}{b}\right)^{2/3} = 1 \)
Answer: From the first equation: \[ x = a \cos^3\theta \implies \frac{x}{a} = \cos^3\theta \] Taking the power of \( 2/3 \) on both sides: \[ \left(\frac{x}{a}\right)^{2/3} = \left(\cos^3\theta\right)^{2/3} = \cos^2\theta \] (Equation 1) From the second equation: \[ y = b \sin^3\theta \implies \frac{y}{b} = \sin^3\theta \] Taking the power of \( 2/3 \) on both sides: \[ \left(\frac{y}{b}\right)^{2/3} = \left(\sin^3\theta\right)^{2/3} = \sin^2\theta \] (Equation 2) Now, add Equation 1 and Equation 2: \[ \left(\frac{x}{a}\right)^{2/3} + \left(\frac{y}{b}\right)^{2/3} = \cos^2\theta + \sin^2\theta \] Using the identity \( \cos^2\theta + \sin^2\theta = 1 \): \[ \left(\frac{x}{a}\right)^{2/3} + \left(\frac{y}{b}\right)^{2/3} = 1 \] Hence proved.
In simple words: Isolate the cosine and sine cubed terms, then raise both sides to the power of \( 2/3 \). Adding the resulting equations gives \( \cos^2\theta + \sin^2\theta \), which equals 1.

Exam Tip: Raising fractional powers like \( 2/3 \) to cubed variables is a simple trick to eliminate the cubes and obtain standard quadratic terms that fit Pythagorean identities.