CBSE Class 10 Mathematics Trigonometry Assignment Set 29

Previous Year Problems (CBSE Board): Part 1

Question 1. If \( 0^\circ < x < 90^\circ \) and \( 2\sin^2x = \dfrac{1}{2} \), then the value of x is [CBSE 2013]
(A) 90°   (B) 30°   (C) 15°   (D) 60°
Answer:
Ans. (B) 30°
\( \sin^2x = \dfrac{1}{4} \Rightarrow \sin x = \dfrac{1}{2} \Rightarrow x = 30^\circ \).
In simple words: Take the positive root since x is acute.

Teacher's Note:
1. sin 30° = 1/2.

Question 2. If \( \tan\theta = \dfrac{1}{\sqrt{7}} \), then the value of \( \dfrac{\operatorname{cosec}^2\theta - \sec^2\theta}{\operatorname{cosec}^2\theta + \sec^2\theta} \) is [CBSE 2013]
(A) \( \dfrac{1}{5} \)   (B) \( \dfrac{3}{4} \)   (C) \( \dfrac{6}{4} \)   (D) \( \dfrac{4}{\sqrt{7}} \)
Answer:
Ans. (B) \( \dfrac{3}{4} \)
\( \cot^2\theta = 7 \), so \( \operatorname{cosec}^2\theta = 1 + 7 = 8 \) and \( \sec^2\theta = 1 + \dfrac{1}{7} = \dfrac{8}{7} \).
The value is \( \dfrac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \dfrac{7 - 1}{7 + 1} = \dfrac{3}{4} \).
In simple words: Find cosec² θ and sec² θ from tan θ.

Teacher's Note:
1. Factor out 8 and cancel.

Question 3. If \( \cot\theta + \dfrac{1}{\cot\theta} = 2 \), then the value of \( \cot^2\theta + \dfrac{1}{\cot^2\theta} \) is [CBSE 2013]
(A) −1   (B) 1   (C) 2   (D) −2
Answer:
Ans. (C) 2
Squaring: \( \cot^2\theta + 2 + \dfrac{1}{\cot^2\theta} = 4 \Rightarrow \cot^2\theta + \dfrac{1}{\cot^2\theta} = 2 \).
In simple words: Square the given equation.

Teacher's Note:
1. The middle term of the square is 2.

Question 4. The value of \( \operatorname{cosec}^2 30^\circ\sin^2 45^\circ - \sec^2 60^\circ \) is [CBSE 2013]
(A) −2   (B) 2   (C) −1   (D) 1
Answer:
Ans. (A) −2
\( \operatorname{cosec}^2 30^\circ = 4 \), \( \sin^2 45^\circ = \dfrac{1}{2} \), \( \sec^2 60^\circ = 4 \). The value is \( 4 \cdot \dfrac{1}{2} - 4 = -2 \).
In simple words: Substitute table values.

Teacher's Note:
1. Square each value before multiplying.

Question 5. Without using trigonometric tables, find the value of \( \dfrac{\cos 70^\circ}{\sin 20^\circ} + \cos 57^\circ\operatorname{cosec}33^\circ - 2\cos 60^\circ \). [2 Marks, CBSE 2013] (Options: (A) −1 (B) 1 (C) −2 (D) 2)
Answer:
\( \cos 70^\circ = \sin 20^\circ \) and \( \cos 57^\circ = \sin 33^\circ \). The value is \( 1 + 1 - 2 \cdot \dfrac{1}{2} = 1 \).
Answer: (B) 1.
OR If A, B, C are interior angles of \( \triangle ABC \), show that \( \cos\left(\dfrac{B + C}{2}\right) = \sin\dfrac{A}{2} \): \( \dfrac{B + C}{2} = 90^\circ - \dfrac{A}{2} \), so \( \cos\left(90^\circ - \dfrac{A}{2}\right) = \sin\dfrac{A}{2} \).
In simple words: Use complementary-angle ratios.

Teacher's Note:
1. cos(90° − x) = sin x.

Question 6. Prove that \( \dfrac{\cos\theta}{\operatorname{cosec}\theta + 1} + \dfrac{\cos\theta}{\operatorname{cosec}\theta - 1} = 2\tan\theta \). [3 Marks, CBSE 2013]
Answer:
LHS \( = \cos\theta \cdot \dfrac{2\operatorname{cosec}\theta}{\operatorname{cosec}^2\theta - 1} = \dfrac{2\cos\theta\operatorname{cosec}\theta}{\cot^2\theta} = \dfrac{2\cos\theta}{\sin\theta} \cdot \dfrac{\sin^2\theta}{\cos^2\theta} = 2\tan\theta \).
In simple words: Combine the fractions.

Teacher's Note:
1. cosec² θ − 1 = cot² θ.

Question 7. Prove that \( \sqrt{\dfrac{1 + \sin\theta}{1 - \sin\theta}} + \sqrt{\dfrac{1 - \sin\theta}{1 + \sin\theta}} = 2\sec\theta \). [3 Marks, CBSE 2013, 2014]
Answer:
\( \sqrt{\dfrac{(1 + \sin\theta)^2}{\cos^2\theta}} + \sqrt{\dfrac{(1 - \sin\theta)^2}{\cos^2\theta}} = \dfrac{1 + \sin\theta}{\cos\theta} + \dfrac{1 - \sin\theta}{\cos\theta} = \dfrac{2}{\cos\theta} = 2\sec\theta \).
In simple words: Multiply inside each root by the numerator.

Teacher's Note:
1. 1 − sin² θ = cos² θ.

Question 8. Without using trigonometric tables, evaluate \( \dfrac{\sec 37^\circ}{\operatorname{cosec}53^\circ} + 2\cot 15^\circ\cot 25^\circ\cot 45^\circ\cot 75^\circ\cot 65^\circ - 3(\sin^2 18^\circ + \sin^2 72^\circ) \). [CBSE 2013]
Answer:
\( \sec 37^\circ = \operatorname{cosec}53^\circ \) so the first term is 1. \( \cot 15^\circ\cot 75^\circ = 1 \), \( \cot 25^\circ\cot 65^\circ = 1 \), \( \cot 45^\circ = 1 \), so the second term is 2. \( \sin^2 72^\circ = \cos^2 18^\circ \), so the bracket is 1.
The value is \( 1 + 2 - 3 = \) 0.
OR Prove \( \dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \sec\theta\operatorname{cosec}\theta \): LHS \( = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} - \dfrac{\cos^2\theta}{\sin\theta(\sin\theta - \cos\theta)} = \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)} = \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = 1 + \sec\theta\operatorname{cosec}\theta \).
In simple words: Pair up the complementary angles.

Teacher's Note:
1. cot(90° − x) = tan x.

Question 9. If \( x = 3\sec^2\theta - 1 \), \( y = \tan^2\theta - 2 \), then \( x - 3y \) is equal to [2 Marks, CBSE 2014]
(A) 3   (B) 4   (C) 5   (D) 8
Answer:
Ans. (D) 8
\( x - 3y = 3\sec^2\theta - 1 - 3\tan^2\theta + 6 = 3(\sec^2\theta - \tan^2\theta) + 5 = 3 + 5 = 8 \).
In simple words: Group the sec² and tan² terms.

Teacher's Note:
1. sec² θ − tan² θ = 1.

Question 10. If \( \tan\theta = \dfrac{2}{3} \), then the value of \( \dfrac{(2 + 2\sec\theta)(1 - \sec\theta)}{(2 + 2\operatorname{cosec}\theta)(1 - \operatorname{cosec}\theta)} \) is [2 Marks, CBSE 2014]
(A) \( \dfrac{81}{16} \)   (B) \( \dfrac{16}{81} \)   (C) \( \dfrac{75}{16} \)   (D) \( \dfrac{77}{16} \)
Answer:
Ans. (B) \( \dfrac{16}{81} \)
Numerator \( = 2(1 + \sec\theta)(1 - \sec\theta) = 2(1 - \sec^2\theta) = -2\tan^2\theta \). Denominator \( = 2(1 - \operatorname{cosec}^2\theta) = -2\cot^2\theta \).
The ratio is \( \dfrac{\tan^2\theta}{\cot^2\theta} = \tan^4\theta = \dfrac{16}{81} \).
In simple words: Use (1 + a)(1 − a) = 1 − a².

Teacher's Note:
1. The ratio simplifies to tan⁴ θ.

Question 11. If \( \sin\theta + \sin^2\theta = 1 \), then the value of \( \cos^2\theta + \cos^4\theta \) is [2 Marks, CBSE 2014]
(A) 2   (B) 1   (C) −2   (D) −1
Answer:
Ans. (B) 1
\( \sin\theta = 1 - \sin^2\theta = \cos^2\theta \). So \( \cos^2\theta + \cos^4\theta = \sin\theta + \sin^2\theta = 1 \).
In simple words: The given relation says sin θ = cos² θ.

Teacher's Note:
1. Replace cos² θ by sin θ.

Question 12. Prove that \( \sec^4\theta - \sec^2\theta = \tan^4\theta + \tan^2\theta \). [CBSE 2014]
Answer:
LHS \( = \sec^2\theta(\sec^2\theta - 1) = (1 + \tan^2\theta)\tan^2\theta = \tan^2\theta + \tan^4\theta = \) RHS.
In simple words: Factor sec² θ out.

Teacher's Note:
1. sec² θ − 1 = tan² θ.

Question 13. OR: Find acute angles A and B, if \( \sin(A + 2B) = \dfrac{\sqrt{3}}{2} \) and \( \cos(A + 4B) = 0 \), \( A > B \). [CBSE 2014]
Answer:
\( A + 2B = 60^\circ \) and \( A + 4B = 90^\circ \). Subtracting: \( 2B = 30^\circ \Rightarrow B = 15^\circ \) and \( A = 30^\circ \).
(The other value \( A + 2B = 120^\circ \) would give \( 2B = -30^\circ \), which is not possible.) A = 30°, B = 15°.
In simple words: Read the angles from sin 60° and cos 90°.

Teacher's Note:
1. A > B is satisfied: 30° > 15°.

Question 14. Prove that \( \tan^2A + \cot^2A = \sec^2A\operatorname{cosec}^2A - 2 \). [CBSE 2014]
Answer:
RHS \( = \dfrac{1}{\sin^2A\cos^2A} - 2 = \dfrac{(\sin^2A + \cos^2A)^2 - 2\sin^2A\cos^2A}{\sin^2A\cos^2A} = \dfrac{\sin^4A + \cos^4A}{\sin^2A\cos^2A} = \tan^2A + \cot^2A \).
In simple words: Write 1 as (sin² + cos²)².

Teacher's Note:
1. sin⁴/(sin² cos²) = tan² A.

Question 15. Prove that \( \sqrt{\dfrac{\sec\theta - 1}{\sec\theta + 1}} + \sqrt{\dfrac{\sec\theta + 1}{\sec\theta - 1}} = 2\operatorname{cosec}\theta \). [3 Marks, CBSE 2014]
Answer:
Rationalise: \( \dfrac{\sec\theta - 1}{\sqrt{\sec^2\theta - 1}} + \dfrac{\sec\theta + 1}{\sqrt{\sec^2\theta - 1}} = \dfrac{2\sec\theta}{\tan\theta} = \dfrac{2}{\sin\theta} = 2\operatorname{cosec}\theta \).
In simple words: Multiply each root's inside by the numerator.

Teacher's Note:
1. sec² θ − 1 = tan² θ.

Question 16. Prove that \( \dfrac{\cot\theta - 1 + \operatorname{cosec}\theta}{\cot\theta + 1 - \operatorname{cosec}\theta} = \dfrac{1}{\operatorname{cosec}\theta - \cot\theta} \). [3 Marks, CBSE 2014] OR If \( \tan\theta + \sin\theta = m \) and \( \tan\theta - \sin\theta = n \), show that \( (m^2 - n^2)^2 = 16mn \).
Answer:
Write \( 1 = \operatorname{cosec}^2\theta - \cot^2\theta \) in the numerator: numerator \( = (\cot\theta + \operatorname{cosec}\theta) - (\operatorname{cosec}\theta - \cot\theta)(\operatorname{cosec}\theta + \cot\theta) = (\cot\theta + \operatorname{cosec}\theta)(1 - \operatorname{cosec}\theta + \cot\theta) \).
LHS \( = \cot\theta + \operatorname{cosec}\theta = \dfrac{1}{\operatorname{cosec}\theta - \cot\theta} = \) RHS.
OR: \( m + n = 2\tan\theta \), \( m - n = 2\sin\theta \), so \( m^2 - n^2 = 4\tan\theta\sin\theta \) and \( (m^2 - n^2)^2 = 16\tan^2\theta\sin^2\theta \). Also \( mn = \tan^2\theta - \sin^2\theta = \tan^2\theta(1 - \cos^2\theta) = \tan^2\theta\sin^2\theta \). Hence \( (m^2 - n^2)^2 = 16mn \).
In simple words: Use 1 = cosec² − cot² in the numerator.

Teacher's Note:
1. The OR part uses tan² − sin² = tan² sin².

Question 17. If \( x = \cot A + \cos A \) and \( y = \cot A - \cos A \), show that \( x^2 - y^2 = 4\sqrt{xy} \). [3 Marks, CBSE 2014]
Answer:
\( x^2 - y^2 = (x + y)(x - y) = (2\cot A)(2\cos A) = 4\cot A\cos A \).
\( xy = \cot^2A - \cos^2A = \cos^2A\left(\dfrac{1}{\sin^2A} - 1\right) = \dfrac{\cos^4A}{\sin^2A} \), so \( \sqrt{xy} = \dfrac{\cos^2A}{\sin A} = \cot A\cos A \) (A acute).
Hence \( x^2 - y^2 = 4\sqrt{xy} \).
In simple words: Factor x² − y² and simplify xy.

Teacher's Note:
1. Take the positive square root for an acute angle.

Question 18. Two poles of height 10 m and 15 m stand vertically on a plane ground. If the distance between their feet is \( 5\sqrt{3} \) m, find the distance between their tops. [3 Marks, CBSE 2014]
Answer:
The difference in heights is 5 m. The tops and the feet form a right triangle with legs \( 5\sqrt{3} \) and 5 m.
Distance \( = \sqrt{(5\sqrt{3})^2 + 5^2} = \sqrt{75 + 25} = \) 10 m.
In simple words: Draw a horizontal line from the shorter top.

Teacher's Note:
1. The triangle has angle 30° at the shorter top.

Question 19. Prove that \( \sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \). [3 Marks, CBSE 2014]
Answer:
LHS \( = \sqrt{\dfrac{(1 + \sin A)^2}{1 - \sin^2A}} = \dfrac{1 + \sin A}{\cos A} = \sec A + \tan A \).
In simple words: Multiply inside the root by 1 + sin A.

Teacher's Note:
1. 1 − sin² A = cos² A.

Question 20. If \( \sin\alpha = \dfrac{1}{2} \), then the value of \( 4\cos^3\alpha - 3\cos\alpha \) is [2 Marks, CBSE 2015]
(A) 0   (B) 1   (C) −1   (D) 2
Answer:
Ans. (A) 0
\( \alpha = 30^\circ \): \( 4\left(\dfrac{\sqrt{3}}{2}\right)^3 - 3 \cdot \dfrac{\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2} = 0 \).
In simple words: Find α first.

Teacher's Note:
1. α = 30° and cos 30° = √3/2.

Question 21. If \( \cos 2\theta = \sin(\theta - 12^\circ) \), where 2θ and (θ − 12°) are both acute angles, then the value of θ is [2 Marks, CBSE 2015]
(A) 24°   (B) 28°   (C) 32°   (D) 34°
Answer:
Ans. (D) 34°
\( \cos 2\theta = \sin(90^\circ - 2\theta) \), so \( 90^\circ - 2\theta = \theta - 12^\circ \Rightarrow 3\theta = 102^\circ \Rightarrow \theta = 34^\circ \).
In simple words: Convert cos to sin.

Teacher's Note:
1. Both angles being acute lets us equate them.