RS Aggarwal Class 10 Mathematics Solutions Chapter 5 Trigonometric Ratios

Question 1. If sin θ = √3/2, find all the other trigonometric ratios.
Answer: Start by drawing a right triangle ABC with the right angle at B and ∠C = θ. Since sin θ = perpendicular/hypotenuse = AB/AC = √3/2, we can set AB = √3k and AC = 2k (where k is a positive constant). Using the Pythagorean theorem:

AC² = AB² + BC²

\( (2k)^2 = (\sqrt{3}k)^2 + BC^2 \)

\( 4k^2 = 3k^2 + BC^2 \)

\( BC = k \)

Now we can determine all other trigonometric ratios:

\( \cos \theta = \frac{BC}{AC} = \frac{k}{2k} = \frac{1}{2} \)

\( \tan \theta = \frac{AB}{BC} = \frac{\sqrt{3}k}{k} = \sqrt{3} \)

\( \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\sqrt{3}} \)

\( \csc \theta = \frac{1}{\sin \theta} = \frac{2}{\sqrt{3}} \)

\( \sec \theta = \frac{1}{\cos \theta} = 2 \)
In simple words: When you know one trigonometric ratio, you can find all the others by setting up a right triangle and using the Pythagorean theorem to locate the missing side.

Exam Tip: Always express ratios in simplified form and verify using the Pythagorean theorem before finalizing your answer.

Question 2. If cos θ = 7/25, find all the other trigonometric ratios.
Answer: Construct a right triangle ABC with the right angle at B and ∠C = θ. Since cos θ = base/hypotenuse = BC/AC = 7/25, let BC = 7k and AC = 25k (where k is a positive constant). Applying the Pythagorean theorem:

AC² = AB² + BC²

\( (25k)^2 = AB^2 + (7k)^2 \)

\( AB^2 = 625k^2 - 49k^2 = 576k^2 \)

\( AB = 24k \)

The remaining trigonometric ratios are:

\( \sin \theta = \frac{AB}{AC} = \frac{24k}{25k} = \frac{24}{25} \)

\( \tan \theta = \frac{AB}{BC} = \frac{24k}{7k} = \frac{24}{7} \)

\( \cot \theta = \frac{1}{\tan \theta} = \frac{7}{24} \)

\( \csc \theta = \frac{1}{\sin \theta} = \frac{25}{24} \)

\( \sec \theta = \frac{1}{\cos \theta} = \frac{25}{7} \)
In simple words: Begin with the given ratio, use it to label two sides of a right triangle, then apply the Pythagorean theorem to locate the third side before calculating the remaining ratios.

Exam Tip: Keep track of which ratio (sine, cosine, tangent) is given so you label the correct sides of your triangle.

Question 3. If tan θ = 15/8, find all the other trigonometric ratios.
Answer: Draw a right triangle ABC with the right angle at B and ∠C = θ. Given tan θ = perpendicular/base = AB/BC = 15/8, set AB = 15k and BC = 8k (where k is a positive constant). By the Pythagorean theorem:

AC² = AB² + BC²

\( AC^2 = (15k)^2 + (8k)^2 = 225k^2 + 64k^2 = 289k^2 \)

\( AC = 17k \)

The remaining trigonometric ratios become:

\( \sin \theta = \frac{AB}{AC} = \frac{15k}{17k} = \frac{15}{17} \)

\( \cos \theta = \frac{BC}{AC} = \frac{8k}{17k} = \frac{8}{17} \)

\( \cot \theta = \frac{1}{\tan \theta} = \frac{8}{15} \)

\( \csc \theta = \frac{1}{\sin \theta} = \frac{17}{15} \)

\( \sec \theta = \frac{1}{\cos \theta} = \frac{17}{8} \)
In simple words: When tangent is given, the two numbers in the fraction go directly to the perpendicular and base of your triangle, then you find the hypotenuse using the Pythagorean theorem.

Exam Tip: Remember that tan θ = opposite/adjacent, so assign those two values carefully before using the Pythagorean theorem.

Question 4. If cot θ = 2, find all the other trigonometric ratios.
Answer: Construct a right triangle ABC with the right angle at B and ∠C = θ. Given cot θ = base/perpendicular = BC/AB = 2, set BC = 2k and AB = k (where k is a positive constant). Using the Pythagorean theorem:

AC² = AB² + BC²

\( AC^2 = (k)^2 + (2k)^2 = k^2 + 4k^2 = 5k^2 \)

\( AC = \sqrt{5}k \)

The remaining trigonometric ratios are:

\( \sin \theta = \frac{AB}{AC} = \frac{k}{\sqrt{5}k} = \frac{1}{\sqrt{5}} \)

\( \cos \theta = \frac{BC}{AC} = \frac{2k}{\sqrt{5}k} = \frac{2}{\sqrt{5}} \)

\( \tan \theta = \frac{1}{\cot \theta} = \frac{1}{2} \)

\( \csc \theta = \frac{1}{\sin \theta} = \sqrt{5} \)

\( \sec \theta = \frac{1}{\cos \theta} = \frac{\sqrt{5}}{2} \)
In simple words: Cotangent is the reciprocal of tangent, so the numbers in cot θ are the base and perpendicular—find the hypotenuse next to get all six ratios.

Exam Tip: Always rationalize denominators in your final answers for a clean, simplified form.

Question 5. If cosec θ = √10, find all the other trigonometric ratios.
Answer: Create a right triangle ABC with the right angle at B and ∠C = θ. Since cosec θ = hypotenuse/perpendicular = AC/AB = √10/1, set AC = √10 k and AB = k (where k is a positive constant). By the Pythagorean theorem:

AC² = AB² + BC²

\( (\sqrt{10}k)^2 = (k)^2 + BC^2 \)

\( 10k^2 = k^2 + BC^2 \)

\( BC^2 = 9k^2 \)

\( BC = 3k \)

The remaining trigonometric ratios become:

\( \sin \theta = \frac{1}{\csc \theta} = \frac{1}{\sqrt{10}} \)

\( \cos \theta = \frac{BC}{AC} = \frac{3k}{\sqrt{10}k} = \frac{3}{\sqrt{10}} \)

\( \tan \theta = \frac{AB}{BC} = \frac{k}{3k} = \frac{1}{3} \)

\( \cot \theta = \frac{1}{\tan \theta} = 3 \)

\( \sec \theta = \frac{1}{\cos \theta} = \frac{\sqrt{10}}{3} \)
In simple words: Cosecant is the hypotenuse divided by the perpendicular, so use these to set up the triangle, then calculate the third side to find the rest.

Exam Tip: Cosecant and sine are reciprocals, as are secant and cosine—use this relationship to check your work quickly.

Question 6. If sin θ = (a² - b²)/(a² + b²), find all the other trigonometric ratios.
Answer: Start with sin θ = (a² - b²)/(a² + b²). To locate the other ratios, use the identity cos² θ = 1 - sin² θ:

\( \cos^2 \theta = 1 - \left(\frac{a^2 - b^2}{a^2 + b^2}\right)^2 = \frac{(a^2 + b^2)^2 - (a^2 - b^2)^2}{(a^2 + b^2)^2} \)

Expand using difference of squares:

\( = \frac{[(a^2 + b^2) - (a^2 - b^2)][(a^2 + b^2) + (a^2 - b^2)]}{(a^2 + b^2)^2} = \frac{[2b^2][2a^2]}{(a^2 + b^2)^2} = \frac{4a^2b^2}{(a^2 + b^2)^2} \)

So:

\( \cos \theta = \frac{2ab}{a^2 + b^2} \)

\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{a^2 - b^2}{2ab} \)

\( \csc \theta = \frac{1}{\sin \theta} = \frac{a^2 + b^2}{a^2 - b^2} \)

\( \sec \theta = \frac{1}{\cos \theta} = \frac{a^2 + b^2}{2ab} \)

\( \cot \theta = \frac{1}{\tan \theta} = \frac{2ab}{a^2 - b^2} \)
In simple words: When sine involves algebraic expressions, apply the Pythagorean identity to locate cosine, then compute the remaining four ratios using reciprocals and division.

Exam Tip: Factor carefully when dealing with algebraic expressions—the difference of squares formula simplifies the work significantly.

Question 7. If 15 cot A = 8, find sin A, cos A, and sec A.
Answer: From 15 cot A = 8, we get cot A = 8/15. Using the identity cosec² A = 1 + cot² A:

\( \csc^2 A = 1 + \left(\frac{8}{15}\right)^2 = 1 + \frac{64}{225} = \frac{289}{225} \)

\( \csc A = \frac{17}{15} \)

Therefore:

\( \sin A = \frac{1}{\csc A} = \frac{15}{17} \)

To get cos A, use cos² A = 1 - sin² A:

\( \cos^2 A = 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{64}{289} \)

\( \cos A = \frac{8}{17} \)

\( \sec A = \frac{1}{\cos A} = \frac{17}{8} \)
In simple words: When cotangent is given, use the Pythagorean identity to locate cosecant and sine, then apply another identity to get cosine and its reciprocal.

Exam Tip: Always simplify the given equation first, then systematically apply one identity at a time to avoid errors.

Question 8. If sin A = 9/41, find cos A and tan A.
Answer: Given sin A = 9/41, apply cos² A = 1 - sin² A:

\( \cos^2 A = 1 - \left(\frac{9}{41}\right)^2 = 1 - \frac{81}{1681} = \frac{1600}{1681} \)

\( \cos A = \frac{40}{41} \)

Now compute tan A:

\( \tan A = \frac{\sin A}{\cos A} = \frac{\frac{9}{41}}{\frac{40}{41}} = \frac{9}{40} \)
In simple words: Given sine, use the Pythagorean identity to find cosine, then divide sine by cosine to get tangent.

Exam Tip: When computing cosine from sine, always take the positive square root (unless explicitly told the angle is obtuse).

Question 9. If cos θ = 0.6, prove that 5 sin θ - 3 tan θ = 0.
Answer: Convert cos θ = 0.6 = 6/10 = 3/5. Build a right triangle ABC with the right angle at B and ∠C = θ. Set BC = 3k and AC = 5k (where k is a positive constant). By the Pythagorean theorem:

AC² = AB² + BC²

\( (5k)^2 = AB^2 + (3k)^2 \)

\( AB^2 = 25k^2 - 9k^2 = 16k^2 \)

\( AB = 4k \)

Now compute sin θ and tan θ:

\( \sin \theta = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5} \)

\( \tan \theta = \frac{AB}{BC} = \frac{4k}{3k} = \frac{4}{3} \)

Substitute into the left side:

\( 5 \sin \theta - 3 \tan \theta = 5 \left(\frac{4}{5}\right) - 3 \left(\frac{4}{3}\right) = 4 - 4 = 0 \)

Thus LHS = RHS, which proves the identity.
In simple words: Set up a triangle from the given cosine value, calculate sine and tangent, then substitute them into the expression to confirm it equals zero.

Exam Tip: Always work with exact fractions rather than decimals when possible—it simplifies algebra and reduces rounding errors.

Question 10. If cosec θ = 2, prove that cot θ + sin θ/(1 + cos θ) = 2.
Answer: Given cosec θ = 2, we have sin θ = 1/cosec θ = 1/2. Set up a right triangle ABC with the right angle at B and ∠C = θ. If AB = k and AC = 2k, then by the Pythagorean theorem:

AC² = AB² + BC²

\( (2k)^2 = (k)^2 + BC^2 \)

\( BC^2 = 3k^2 \)

\( BC = \sqrt{3}k \)

Compute the other ratios:

\( \cos \theta = \frac{BC}{AC} = \frac{\sqrt{3}k}{2k} = \frac{\sqrt{3}}{2} \)

\( \tan \theta = \frac{AB}{BC} = \frac{k}{\sqrt{3}k} = \frac{1}{\sqrt{3}} \)

\( \cot \theta = \frac{1}{\tan \theta} = \sqrt{3} \)

Now evaluate the left side:

\( \cot \theta + \frac{\sin \theta}{1 + \cos \theta} = \sqrt{3} + \frac{\frac{1}{2}}{1 + \frac{\sqrt{3}}{2}} = \sqrt{3} + \frac{\frac{1}{2}}{\frac{2 + \sqrt{3}}{2}} = \sqrt{3} + \frac{1}{2 + \sqrt{3}} \)

\( = \frac{\sqrt{3}(2 + \sqrt{3}) + 1}{2 + \sqrt{3}} = \frac{2\sqrt{3} + 3 + 1}{2 + \sqrt{3}} = \frac{2(\sqrt{3} + 2)}{2 + \sqrt{3}} = 2 \)

Thus LHS = RHS = 2, completing the proof.
In simple words: Build a triangle from the given cosecant, find all six trigonometric ratios, substitute them into the left side, and simplify carefully to match the right side.

Exam Tip: When the expression contains a fraction within the fraction, simplify from the inside out to avoid mistakes.

Question 11. If tan θ = 1/√7, prove that (cosec² θ - sec² θ)/(cosec² θ + sec² θ) = 3/4.
Answer: Create a right triangle ABC with the right angle at B and ∠C = θ. From tan θ = 1/√7, set AB = k and BC = √7 k (where k is a positive constant). Use the Pythagorean theorem:

AC² = AB² + BC²

\( AC^2 = (k)^2 + (\sqrt{7}k)^2 = k^2 + 7k^2 = 8k^2 \)

\( AC = 2\sqrt{2}k \)

Now compute the required reciprocal ratios:

\( \sin \theta = \frac{AB}{AC} = \frac{k}{2\sqrt{2}k} = \frac{1}{2\sqrt{2}} \)

\( \cos \theta = \frac{BC}{AC} = \frac{\sqrt{7}k}{2\sqrt{2}k} = \frac{\sqrt{7}}{2\sqrt{2}} \)

\( \csc \theta = \frac{1}{\sin \theta} = 2\sqrt{2} \)

\( \sec \theta = \frac{1}{\cos \theta} = \frac{2\sqrt{2}}{\sqrt{7}} \)

Substitute into the expression:

\( \frac{\csc^2 \theta - \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta} = \frac{(2\sqrt{2})^2 - \left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^2}{(2\sqrt{2})^2 + \left(\frac{2\sqrt{2}}{\sqrt{7}}\right)^2} = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{56 - 8}{7}}{\frac{56 + 8}{7}} = \frac{48}{64} = \frac{3}{4} \)

Thus LHS = RHS, which proves the identity.
In simple words: From tan θ, create a triangle and find the hypotenuse using the Pythagorean theorem, then calculate cosecant and secant before substituting into the expression.

Exam Tip: Always square the values before adding or subtracting them in the numerator and denominator.

Question 12. If tan θ = 20/21, prove that (1 - sin θ + cos θ)/(1 + sin θ + cos θ) = 3/7.
Answer: Construct a right triangle ABC with the right angle at B and ∠C = θ. Given tan θ = 20/21, set AB = 20k and BC = 21k (where k is a positive constant). Apply the Pythagorean theorem:

AC² = AB² + BC²

\( AC^2 = (20k)^2 + (21k)^2 = 400k^2 + 441k^2 = 841k^2 \)

\( AC = 29k \)

Determine sin θ and cos θ:

\( \sin \theta = \frac{AB}{AC} = \frac{20k}{29k} = \frac{20}{29} \)

\( \cos \theta = \frac{BC}{AC} = \frac{21k}{29k} = \frac{21}{29} \)

Substitute into the left side:

\( \text{LHS} = \frac{1 - \sin \theta + \cos \theta}{1 + \sin \theta + \cos \theta} = \frac{1 - \frac{20}{29} + \frac{21}{29}}{1 + \frac{20}{29} + \frac{21}{29}} = \frac{\frac{29 - 20 + 21}{29}}{\frac{29 + 20 + 21}{29}} = \frac{30}{70} = \frac{3}{7} \)

Therefore LHS = RHS, confirming the identity.
In simple words: Extract sine and cosine from the given tangent value using a right triangle and the Pythagorean theorem, then substitute them directly into the fraction.

Exam Tip: Work with a common denominator when adding or subtracting fractions in the numerator and denominator separately.

Question 13. If sec θ = 5/4, prove that (sin θ - 2 cos θ)/(tan θ - cot θ) = 12/7.
Answer: Given sec θ = 5/4, we have cos θ = 1/sec θ = 4/5. Use cos² θ = 1 - sin² θ:

\( \sin^2 \theta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \)

\( \sin \theta = \frac{3}{5} \)

Compute tan θ and cot θ:

\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)

\( \cot \theta = \frac{1}{\tan \theta} = \frac{4}{3} \)

Evaluate the left side:

\( \text{LHS} = \frac{\sin \theta - 2 \cos \theta}{\tan \theta - \cot \theta} = \frac{\frac{3}{5} - 2 \cdot \frac{4}{5}}{\frac{3}{4} - \frac{4}{3}} = \frac{\frac{3}{5} - \frac{8}{5}}{\frac{9 - 16}{12}} = \frac{\frac{-5}{5}}{\frac{-7}{12}} = \frac{-1}{\frac{-7}{12}} = \frac{12}{7} \)

Thus LHS = RHS, which completes the proof.
In simple words: From the given secant, find cosine, then use an identity to locate sine, calculate tangent and cotangent, and substitute all values into the fraction.

Exam Tip: When subtracting fractions, find a common denominator first to simplify the algebra.

Question 14. Prove that √[(sec θ - cosec θ)/(sec θ + cosec θ)] = √[(1 - cos θ)/(1 + cos θ)].
Answer: Begin with the left side and express sec θ and cosec θ in terms of sine and cosine:

\( \text{LHS} = \sqrt{\frac{\frac{1}{\cos \theta} - \frac{1}{\sin \theta}}{\frac{1}{\cos \theta} + \frac{1}{\sin \theta}}} = \sqrt{\frac{\frac{\sin \theta - \cos \theta}{\sin \theta \cos \theta}}{\frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}}} = \sqrt{\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}} \)

Rewrite the numerator and denominator by factoring out sine:

\( = \sqrt{\frac{\frac{\sin \theta - \cos \theta}{\sin \theta}}{\frac{\sin \theta + \cos \theta}{\sin \theta}}} = \sqrt{\frac{1 - \frac{\cos \theta}{\sin \theta}}{1 + \frac{\cos \theta}{\sin \theta}}} = \sqrt{\frac{1 - \cot \theta}{1 + \cot \theta}} \)

Alternatively, simplify directly by dividing both numerator and denominator by sine and cosine:

\( = \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = \text{RHS} \)

Thus LHS = RHS, confirming the identity.
In simple words: Convert secant and cosecant to sine and cosine fractions, simplify the complex fraction, and match it to the right side.

Exam Tip: Always express reciprocal ratios in terms of basic sine and cosine to simplify complex algebraic manipulations.

Question 15. Prove that √[(cosec² θ - cot² θ)/(sec² θ - 1)] = cot θ = √(cosec² θ - 1).
Answer: Use the fundamental identities: cosec² θ - cot² θ = 1 and sec² θ - 1 = tan² θ. Substitute:

\( \sqrt{\frac{\cosec^2 \theta - \cot^2 \theta}{\sec^2 \theta - 1}} = \sqrt{\frac{1}{\tan^2 \theta}} = \sqrt{\cot^2 \theta} = \cot \theta \)

For the second equality, use cosec² θ - cot² θ = 1, which gives cot² θ = cosec² θ - 1. Therefore:

\( \cot \theta = \sqrt{\cosec^2 \theta - 1} \)

Thus all three expressions are equal, completing the proof.
In simple words: Apply the Pythagorean identities to simplify the fractions and radicals, then match each form.

Exam Tip: Always recall the three Pythagorean identities: sin² θ + cos² θ = 1, sec² θ - tan² θ = 1, and cosec² θ - cot² θ = 1.

Question 16. If sin θ = a/b, prove that (sec θ + tan θ) = √[(b + a)/(b - a)].
Answer: Start with sec θ + tan θ:

\( \sec \theta + \tan \theta = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin \theta}{\cos \theta} = \frac{1 + \sin \theta}{\sqrt{1 - \sin^2 \theta}} \)

Substitute sin θ = a/b:

\( = \frac{1 + \frac{a}{b}}{\sqrt{1 - \left(\frac{a}{b}\right)^2}} = \frac{\frac{b + a}{b}}{\sqrt{\frac{b^2 - a^2}{b^2}}} = \frac{\frac{b + a}{b}}{\frac{\sqrt{b^2 - a^2}}{b}} = \frac{b + a}{\sqrt{b^2 - a^2}} \)

Factor the denominator:

\( = \frac{b + a}{\sqrt{(b + a)(b - a)}} = \frac{b + a}{\sqrt{b + a}\sqrt{b - a}} = \frac{\sqrt{b + a}}{\sqrt{b - a}} = \sqrt{\frac{b + a}{b - a}} \)

Thus LHS = RHS, proving the identity.
In simple words: Express secant and tangent as a single fraction with cosine in the denominator, then substitute the given sine value and simplify by factoring.

Exam Tip: Always factor square roots when possible to simplify radical expressions.

Question 17. If cos θ = 3/5, prove that (sin θ - cot θ)/(2 tan θ) = 3/160.
Answer: Given cos θ = 3/5, find sin θ using sin² θ = 1 - cos² θ:

\( \sin^2 \theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \)

\( \sin \theta = \frac{4}{5} \)

Compute cot θ and tan θ:

\( \cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)

\( \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \)

Substitute into the left side:

\( \text{LHS} = \frac{\sin \theta - \cot \theta}{2 \tan \theta} = \frac{\frac{4}{5} - \frac{3}{4}}{2 \cdot \frac{4}{3}} = \frac{\frac{16 - 15}{20}}{\frac{8}{3}} = \frac{\frac{1}{20}}{\frac{8}{3}} = \frac{1}{20} \cdot \frac{3}{8} = \frac{3}{160} \)

Thus LHS = RHS, completing the proof.
In simple words: From the given cosine, calculate sine using the Pythagorean identity, then find cotangent and tangent before substituting into the expression.

Exam Tip: Always simplify compound fractions by multiplying by the reciprocal of the denominator.

Question 18. If tan θ = 4/3, prove that (sin θ + cos θ) = 7/5.
Answer: Build a right triangle ABC with the right angle at B and ∠C = θ. From tan θ = 4/3, set BC = 3k and AB = 4k (where k is a positive constant). Use the Pythagorean theorem:

AC² = AB² + BC²

\( AC^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2 \)

\( AC = 5k \)

Calculate sin θ and cos θ:

\( \sin \theta = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5} \)

\( \cos \theta = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5} \)

Add them:

\( \sin \theta + \cos \theta = \frac{4}{5} + \frac{3}{5} = \frac{7}{5} \)

Thus LHS = RHS, proving the identity.
In simple words: Create a triangle from the given tangent value, find the hypotenuse using the Pythagorean theorem, then compute both sine and cosine and add them.

Exam Tip: Always verify that your triangle satisfies the Pythagorean theorem before computing the trigonometric ratios.

Question 19. If tan θ = a/b, prove that (a sin θ - b cos θ)/(a sin θ + b cos θ) = (a² - b²)/(a² + b²).
Answer: To simplify the left side, divide both the numerator and denominator by cos θ:

\( \frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta} = \frac{a \tan \theta - b}{a \tan \theta + b} \)

(since tan θ = sin θ / cos θ)

Substitute tan θ = a/b:

\( = \frac{a \cdot \frac{a}{b} - b}{a \cdot \frac{a}{b} + b} = \frac{\frac{a^2}{b} - b}{\frac{a^2}{b} + b} = \frac{\frac{a^2 - b^2}{b}}{\frac{a^2 + b^2}{b}} = \frac{a^2 - b^2}{a^2 + b^2} \)

Thus LHS = RHS, confirming the identity.
In simple words: Divide the fraction by cosine to convert it into a form involving tangent, then substitute the given value of tan θ and simplify.

Exam Tip: Dividing numerator and denominator by the same function is a powerful technique for introducing a specific trigonometric ratio.

Question 20. If tan θ = 4/3, prove that (sin θ + cos θ) = 7/5.
Answer: Create a right triangle ABC with the right angle at B and ∠C = θ. Given tan θ = 4/3, set AB = 4k and BC = 3k (where k is a positive constant). Use the Pythagorean theorem:

AC² = AB² + BC²

\( AC^2 = (4k)^2 + (3k)^2 = 16k^2 + 9k^2 = 25k^2 \)

\( AC = 5k \)

Determine sin θ and cos θ:

\( \sin \theta = \frac{AB}{AC} = \frac{4k}{5k} = \frac{4}{5} \)

\( \cos \theta = \frac{BC}{AC} = \frac{3k}{5k} = \frac{3}{5} \)

Add the ratios:

\( \sin \theta + \cos \theta = \frac{4}{5} + \frac{3}{5} = \frac{7}{5} \)

Therefore LHS = RHS, completing the proof.
In simple words: Use the given tangent to build a right triangle, apply the Pythagorean theorem to find the hypotenuse, then calculate sine and cosine and add them together.

Exam Tip: Always double-check that the Pythagorean theorem holds for your chosen side lengths before proceeding.

Question 20. Prove that \( \frac{4\cos\theta - \sin\theta}{2\cos\theta + \sin\theta} = \frac{4}{5} \), given the ratio AB:BC = 4:3 in a right triangle.
Answer: Given the ratio AB:BC = 4:3, we can set BC = 3k and AB = 4k for some positive value k. Applying the Pythagorean theorem yields \( AC^2 = AB^2 + BC^2 = 16k^2 + 9k^2 = 25k^2 \), so AC = 5k. This gives us \( \sin\theta = \frac{AB}{AC} = \frac{4}{5} \) and \( \cos\theta = \frac{BC}{AC} = \frac{3}{5} \). Substituting these values into the left side: \( \frac{4\cos\theta - \sin\theta}{2\cos\theta + \sin\theta} = \frac{4 \cdot \frac{3}{5} - \frac{4}{5}}{2 \cdot \frac{3}{5} + \frac{4}{5}} = \frac{\frac{12-4}{5}}{\frac{6+4}{5}} = \frac{8}{10} = \frac{4}{5} \) = RHS. Hence verified.
In simple words: Set up the triangle sides using the given ratio, find the hypotenuse, calculate sin and cos, then plug them into the expression to confirm both sides match.

Exam Tip: Always assign variables to the ratio (like 3k and 4k) rather than fixed numbers - this preserves the proportional structure and makes Pythagorean calculations cleaner.

Question 21. Prove that \( \frac{4\sin\theta - 3\cos\theta}{2\sin\theta + 6\cos\theta} = \frac{1}{3} \), given that \( \cos\theta = \frac{2}{3} \).
Answer: Given \( \cos\theta = \frac{2}{3} \), we first compute \( \sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \). Next, we find \( \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}} \). Dividing the left side by sin θ: \( \frac{4\sin\theta - 3\cos\theta}{2\sin\theta + 6\cos\theta} = \frac{4 - 3\cot\theta}{2 + 6\cot\theta} = \frac{4 - 3 \cdot \frac{2}{3}}{2 + 6 \cdot \frac{2}{3}} = \frac{4 - 2}{2 + 4} = \frac{2}{6} = \frac{1}{3} \) = RHS. Hence verified.
In simple words: Use the given cosine value to find cotangent, divide the entire expression by sin θ to simplify it into cotangent form, then substitute and simplify to reach the target answer.

Exam Tip: When division by sin θ is possible, rewrite the expression in terms of cot θ = cos θ / sin θ to avoid computing sin θ explicitly if only cos θ is given.

Question 22. Prove that \( \frac{(1-\tan^2\theta)}{(1+\tan^2\theta)} = \cos^2\theta - \sin^2\theta \), given that 3 cot θ = 4.
Answer: Given 3 cot θ = 4, we have \( \cot\theta = \frac{4}{3} \). Starting with the left side, we rewrite using \( \tan\theta = \frac{1}{\cot\theta} \): \( \frac{(1-\tan^2\theta)}{(1+\tan^2\theta)} = \frac{(1 - \frac{1}{\cot^2\theta})}{(1 + \frac{1}{\cot^2\theta})} = \frac{\cot^2\theta - 1}{\cot^2\theta + 1} \). Substituting \( \cot\theta = \frac{4}{3} \): \( \frac{(\frac{4}{3})^2 - 1}{(\frac{4}{3})^2 + 1} = \frac{\frac{16}{9} - 1}{\frac{16}{9} + 1} = \frac{\frac{16-9}{9}}{\frac{16+9}{9}} = \frac{7}{25} \). For the right side, \( \cos^2\theta - \sin^2\theta = \frac{\cos^2\theta - \sin^2\theta}{1} = \frac{\cos^2\theta - \sin^2\theta}{\sin^2\theta} \cdot \frac{1}{\text{cosec}^2\theta} = \frac{\cot^2\theta - 1}{\cot^2\theta + 1} = \frac{7}{25} \). LHS = RHS, hence verified.
In simple words: Use the cot value to convert the tangent expression into a cotangent-based form, substitute the given ratio, and confirm the right side yields the same numerical result through a parallel simplification.

Exam Tip: Rewrite tan in terms of cot early on when cot is given - this eliminates the need to find sin and cos separately and keeps all steps algebraic.

Question 23. Prove that \( \frac{3 - 4\sin^2\theta}{4\cos^2\theta - 3} = \frac{3 - \tan^2\theta}{1 - 3\tan^2\theta} \), given that \( \sec\theta = \frac{17}{8} \).
Answer: Given \( \sec\theta = \frac{17}{8} \), we derive \( \cos\theta = \frac{8}{17} \). Setting BC = 8k and AC = 17k in a right triangle, Pythagoras gives \( AB^2 = AC^2 - BC^2 = (17k)^2 - (8k)^2 = 289k^2 - 64k^2 = 225k^2 \), so AB = 15k. Thus \( \sin\theta = \frac{15}{17} \) and \( \tan\theta = \frac{15}{8} \). For the left side: \( \frac{3 - 4\sin^2\theta}{4\cos^2\theta - 3} = \frac{3 - 4(\frac{15}{17})^2}{4(\frac{8}{17})^2 - 3} = \frac{3 - \frac{900}{289}}{\frac{256}{289} - 3} = \frac{\frac{867 - 900}{289}}{\frac{256 - 867}{289}} = \frac{-33}{-611} = \frac{33}{611} \). For the right side: \( \frac{3 - \tan^2\theta}{1 - 3\tan^2\theta} = \frac{3 - (\frac{15}{8})^2}{1 - 3(\frac{15}{8})^2} = \frac{3 - \frac{225}{64}}{1 - \frac{675}{64}} = \frac{\frac{192 - 225}{64}}{\frac{64 - 675}{64}} = \frac{-33}{-611} = \frac{33}{611} \). LHS = RHS, hence proved.
In simple words: Use the secant value to build a right triangle, calculate sin and tan from the sides, then evaluate both sides numerically to confirm they match.

Exam Tip: Construct the triangle first using the reciprocal of sec; this gives you all trig ratios in one step and avoids mistakes from computing them independently.

Question 24. In a composite right triangle configuration with point D above B such that AD = 10 cm and BD = 8 cm, and point C at height 4 cm above B, find (i) sin θ and (ii) cos θ, where θ is the angle at A in triangle ABD looking toward C.
Answer: In triangle ABD with ∠ABD = 90°, applying the Pythagorean theorem: \( AB = \sqrt{AD^2 - BD^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 \text{ cm} \). In triangle ABC with ∠ABC = 90°, applying the Pythagorean theorem: \( AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \text{ cm} \). Therefore: (i) \( \sin\theta = \frac{BC}{AC} = \frac{4}{2\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13} \) (ii) \( \cos\theta = \frac{AB}{AC} = \frac{6}{2\sqrt{13}} = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13} \)
In simple words: Apply Pythagoras twice in sequence - first to find AB from the outer triangle, then to find AC from the inner triangle - then form sine and cosine as ratios of the discovered sides, rationalizing denominators if needed.

Exam Tip: Always rationalize square root denominators in your final answer by multiplying numerator and denominator by the radical - this is the standard form expected in most curricula.

Question 25. In a right triangle ABC with ∠B = 90°, AB = 24 cm, and BC = 7 cm, calculate (i) sin A, (ii) cos A, (iii) sin C, and (iv) cos C.
Answer: Using Pythagoras theorem: \( AC^2 = AB^2 + BC^2 = (24)^2 + (7)^2 = 576 + 49 = 625 \), so AC = 25 cm. For angle A, the base is AB = 24 and the perpendicular is BC = 7. Thus: (i) \( \sin A = \frac{BC}{AC} = \frac{7}{25} \) (ii) \( \cos A = \frac{AB}{AC} = \frac{24}{25} \). For angle C, the base is BC = 7 and the perpendicular is AB = 24. Thus: (iii) \( \sin C = \frac{AB}{AC} = \frac{24}{25} \) (iv) \( \cos C = \frac{BC}{AC} = \frac{7}{25} \)
In simple words: Find the hypotenuse first, then identify which side is opposite and which is adjacent for each angle - opposite over hypotenuse gives sine, adjacent over hypotenuse gives cosine.

Exam Tip: Be careful to swap base and perpendicular when moving from one acute angle to the other in a right triangle - this is a common source of errors.

Question 26. In a right triangle with hypotenuse AB = 29 cm and one leg BC = 21 cm, prove that \( \cos^2\theta - \sin^2\theta = \frac{41}{841} \), where θ is the angle at B.
Answer: Using Pythagoras theorem: \( AC^2 = AB^2 - BC^2 = (29)^2 - (21)^2 = 841 - 441 = 400 \), so AC = 20 units. From the right triangle, \( \sin\theta = \frac{AC}{AB} = \frac{20}{29} \) and \( \cos\theta = \frac{BC}{AB} = \frac{21}{29} \). Computing: \( \cos^2\theta - \sin^2\theta = (\frac{21}{29})^2 - (\frac{20}{29})^2 = \frac{441}{841} - \frac{400}{841} = \frac{41}{841} \). Hence proved.
In simple words: Find the missing side using Pythagoras, determine sin and cos from the triangle's sides, then square each and subtract to verify the given expression equals the target fraction.

Exam Tip: Double-check your identification of which side is opposite and which is adjacent - a single swap will flip the sine and cosine and give the wrong numerical result.

Question 27. In a right triangle ABC with ∠B = 90°, AC = 13 cm, and AB = 12 cm, find (i) cos A, (ii) cosec A, (iii) cos C, and (iv) cosec C.
Answer: Using Pythagoras theorem: \( AC^2 = AB^2 + BC^2 \), so \( BC^2 = AC^2 - AB^2 = (13)^2 - (12)^2 = 169 - 144 = 25 \), giving BC = 5 cm. For angle A, base = AB = 12, perpendicular = BC = 5. Thus: (i) \( \cos A = \frac{AB}{AC} = \frac{12}{13} \) (ii) \( \text{cosec } A = \frac{1}{\sin A} = \frac{AC}{BC} = \frac{13}{5} \). For angle C, base = BC = 5, perpendicular = AB = 12. Thus: (iii) \( \cos C = \frac{BC}{AC} = \frac{5}{13} \) (iv) \( \text{cosec } C = \frac{1}{\sin C} = \frac{AC}{AB} = \frac{13}{12} \)
In simple words: Find BC from Pythagoras, then for each angle identify its base and perpendicular, compute cosine as base over hypotenuse and cosecant as hypotenuse over the opposite side.

Exam Tip: Remember that cosec is the reciprocal of sine, so it equals hypotenuse over opposite - not the other way around. Flip the fraction and you'll get the answer wrong.

Question 28. Prove that \( 3\cos a - 4\cos^3 a = 0 \), given that \( \sin a = \frac{1}{2} \).
Answer: Given \( \sin a = \frac{1}{2} \), we have \( \cos a = \sqrt{1 - \sin^2 a} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \). Computing the left side: \( 3\cos a - 4\cos^3 a = \cos a(3 - 4\cos^2 a) = \frac{\sqrt{3}}{2}[3 - 4(\frac{\sqrt{3}}{2})^2] = \frac{\sqrt{3}}{2}[3 - 4 \cdot \frac{3}{4}] = \frac{\sqrt{3}}{2}[3 - 3] = \frac{\sqrt{3}}{2} \cdot 0 = 0 \). Hence proved.
In simple words: Calculate cos from the given sine using the Pythagorean identity, factor out cos from the expression, substitute the calculated value, and simplify to reach zero.

Exam Tip: Always factor out common terms before substituting - it often reveals that the remaining bracket simplifies to zero or a known value, making the proof much simpler.

Question 29. In a right triangle ABC with ∠B = 90° and \( \tan A = \frac{1}{\sqrt{3}} \), prove (i) \( \sin A \cos C + \cos A \sin C = 1 \) and (ii) \( \cos A \cos C - \sin A \sin C = 0 \).
Answer: Given \( \tan A = \frac{1}{\sqrt{3}} = \frac{BC}{AB} \), let BC = x and AB = x√3. Using Pythagoras: \( AC = \sqrt{AB^2 + BC^2} = \sqrt{(x\sqrt{3})^2 + x^2} = \sqrt{3x^2 + x^2} = \sqrt{4x^2} = 2x \). Thus: \( \sin A = \frac{x}{2x} = \frac{1}{2} \), \( \cos A = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2} \), \( \sin C = \frac{x\sqrt{3}}{2x} = \frac{\sqrt{3}}{2} \), \( \cos C = \frac{x}{2x} = \frac{1}{2} \). (i) \( \sin A \cos C + \cos A \sin C = \frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{1}{4} + \frac{3}{4} = 1 \). (ii) \( \cos A \cos C - \sin A \sin C = \frac{\sqrt{3}}{2} \cdot \frac{1}{2} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \). Both proved.
In simple words: Set the ratio BC:AB from tan, find AC from Pythagoras, then compute all four trig values for both angles, substitute into each expression, and verify the results match the target answers.

Exam Tip: Notice that in a right triangle, if one acute angle has a specific trig ratio, the other acute angle's ratios are often complementary swaps - use this to predict your answers before calculating.

Question 30. In a right triangle ABC with ∠C = 90°, prove that if \( \sin A = \sin B \), then ∠A = ∠B.
Answer: In triangle ABC with ∠C = 90°, we have \( \sin A = \frac{BC}{AB} \) and \( \sin B = \frac{AC}{AB} \). If \( \sin A = \sin B \), then \( \frac{BC}{AB} = \frac{AC}{AB} \), which gives BC = AC. By the converse of the isosceles triangle theorem, angles opposite equal sides are equal, so ∠A = ∠B. Hence proved.
In simple words: Write out the sine formulas for both angles using the triangle's sides, set them equal, cancel the common denominator, and conclude that equal opposite sides mean equal angles.

Exam Tip: Remember that in any triangle, equal sides always have equal opposite angles - this is a fundamental property that often appears in proofs involving sine or other ratio equalities.

Question 31. In a right triangle ABC with ∠C = 90°, prove that if \( \tan A = \tan B \), then ∠A = ∠B.
Answer: In triangle ABC with ∠C = 90°, we have \( \tan A = \frac{BC}{AC} \) and \( \tan B = \frac{AC}{BC} \). If \( \tan A = \tan B \), then \( \frac{BC}{AC} = \frac{AC}{BC} \), which gives \( BC^2 = AC^2 \), so BC = AC. By the converse of the isosceles triangle theorem, angles opposite equal sides are equal, so ∠A = ∠B. Hence proved.
In simple words: Form the tangent equations for both angles, set them equal, cross-multiply to obtain equal sides, then apply the isosceles triangle property to conclude the angles are equal.

Exam Tip: Cross-multiplying tangent ratios is a standard technique - remember that if the products cross to equality, you've found equal sides and therefore equal angles.

Question 32. Prove that if tan A = 1, then \( 2\sin A \cos A = 1 \).
Answer: Given tan A = 1, we have \( \frac{\sin A}{\cos A} = 1 \), which gives sin A = cos A. Subtracting: sin A - cos A = 0. Squaring both sides: \( (\sin A - \cos A)^2 = 0 \), which expands to \( \sin^2 A + \cos^2 A - 2\sin A \cos A = 0 \). Since \( \sin^2 A + \cos^2 A = 1 \), we get \( 1 - 2\sin A \cos A = 0 \), so \( 2\sin A \cos A = 1 \). Hence proved.
In simple words: Use the tangent condition to express sin A in terms of cos A, square the resulting equation to eliminate the equality, apply the Pythagorean identity, and solve for the double-angle product.

Exam Tip: Squaring both sides to eliminate a difference is a powerful algebraic move - just remember to check that your squared equation doesn't introduce extraneous solutions (though it doesn't in this case).

Question 33. In right triangle PQR with ∠Q = 90°, QR = x, and PR = x + 2, find (i) \( (\sqrt{x+1})\cot\phi = \frac{x}{2} \), (ii) \( (\sqrt{x^3+x^2})\tan\theta = \frac{x^2}{2} \), and (iii) \( \cos\theta = \frac{2\sqrt{x+1}}{x+2} \).
Answer: Using Pythagoras theorem: \( PQ = \sqrt{PR^2 - QR^2} = \sqrt{(x+2)^2 - x^2} = \sqrt{x^2 + 4x + 4 - x^2} = \sqrt{4x + 4} = 2\sqrt{x+1} \). (i) \( (\sqrt{x+1})\cot\phi = (\sqrt{x+1}) \times \frac{QR}{PQ} = (\sqrt{x+1}) \times \frac{x}{2\sqrt{x+1}} = \frac{x}{2} \). (ii) \( (\sqrt{x^3+x^2})\tan\theta = (\sqrt{x^2(x+1)}) \times \frac{QR}{PQ} = x\sqrt{x+1} \times \frac{x}{2\sqrt{x+1}} = \frac{x^2}{2} \). (iii) \( \cos\theta = \frac{PQ}{PR} = \frac{2\sqrt{x+1}}{x+2} \)
In simple words: Apply Pythagoras to find the missing side, then for each part multiply or divide by the appropriate trig ratio - the radicals in the coefficient often cancel with those in the trig fraction, leaving simple expressions.

Exam Tip: Watch for radical simplifications by factoring inside the square root and canceling with radicals in the denominator - this is what makes seemingly complex expressions yield clean answers.

Question 34. Prove that \( (\frac{2}{x+y})^2 + (\frac{x-y}{2})^2 - 1 = 0 \), where x = cosec A + cos A and y = cosec A - cos A.
Answer: Substituting x and y: \( \frac{2}{x+y} = \frac{2}{(\text{cosec } A + \cos A) + (\text{cosec } A - \cos A)} = \frac{2}{2\text{cosec } A} = \frac{1}{\text{cosec } A} = \sin A \). And \( \frac{x-y}{2} = \frac{(\text{cosec } A + \cos A) - (\text{cosec } A - \cos A)}{2} = \frac{2\cos A}{2} = \cos A \). Therefore: \( (\frac{2}{x+y})^2 + (\frac{x-y}{2})^2 - 1 = (\sin A)^2 + (\cos A)^2 - 1 = \sin^2 A + \cos^2 A - 1 = 1 - 1 = 0 \). Hence proved.
In simple words: Substitute the definitions of x and y into each fraction, simplify by adding or subtracting numerators, apply trig identities to get sin and cos, then use the fundamental identity to reach zero.

Exam Tip: Always look for algebraic cancellations when expressions have matching terms with opposite signs - addition or subtraction often eliminates complexity dramatically.

Question 35. Prove that \( (\frac{x-y}{x+y})^2 + (\frac{x-y}{2})^2 = 1 \), where x = cot A + cos A and y = cot A - cos A.
Answer: Substituting x and y: \( \frac{x-y}{x+y} = \frac{(\text{cot } A + \cos A) - (\text{cot } A - \cos A)}{(\text{cot } A + \cos A) + (\text{cot } A - \cos A)} = \frac{2\cos A}{2\cot A} = \frac{\cos A}{\cot A} = \frac{\cos A}{\frac{\cos A}{\sin A}} = \sin A \). And \( \frac{x-y}{2} = \frac{2\cos A}{2} = \cos A \). Therefore: \( (\frac{x-y}{x+y})^2 + (\frac{x-y}{2})^2 = (\sin A)^2 + (\cos A)^2 = \sin^2 A + \cos^2 A = 1 \). Hence proved.
In simple words: Substitute x and y definitions into each fraction, simplify differences and sums by canceling matching terms, reduce cotangent to sin and cos form, then apply the fundamental trig identity to get 1.

Exam Tip: When cot appears in a complex fraction, always convert it to cos/sin form and simplify - this usually reveals a familiar trig identity waiting to be applied.