CBSE Class 10 Mathematics Trigonometry Assignment Set 30

Previous Year Problems (CBSE Board): Part 2

Question 1. If \( \tan 2A = \cot(A - 18^\circ) \), where 2A is an acute angle, then the value of A is [CBSE 2015, 2017]
(A) 24°   (B) 12°   (C) 36°   (D) 63°
Answer:
Ans. (C) 36°
\( \cot(A - 18^\circ) = \tan(90^\circ - A + 18^\circ) = \tan(108^\circ - A) \), so \( 2A = 108^\circ - A \Rightarrow 3A = 108^\circ \Rightarrow A = 36^\circ \).
In simple words: Convert cot to tan of the complement.

Teacher's Note:
1. Equal tangents of acute angles have equal angles.

Question 2. If \( \sin 5\theta = \cos 4\theta \), where 5θ and 4θ are acute angles, then the value of θ is [CBSE 2015]
(A) 10°   (B) 100°   (C) 12°   (D) 15°
Answer:
Ans. (A) 10°
\( \sin 5\theta = \sin(90^\circ - 4\theta) \Rightarrow 5\theta = 90^\circ - 4\theta \Rightarrow \theta = 10^\circ \).
In simple words: Convert cos 4θ to a sine.

Teacher's Note:
1. 5θ + 4θ = 90°.

Question 3. If \( \tan\theta = \dfrac{12}{13} \), then the value of \( \dfrac{2\sin\theta\cos\theta}{\cos^2\theta - \sin^2\theta} \) is [CBSE 2015]
(A) \( \dfrac{307}{25} \)   (B) \( \dfrac{312}{25} \)   (C) \( \dfrac{309}{25} \)   (D) \( \dfrac{316}{25} \)
Answer:
Ans. (B) \( \dfrac{312}{25} \)
Divide numerator and denominator by \( \cos^2\theta \): \( \dfrac{2\tan\theta}{1 - \tan^2\theta} = \dfrac{\frac{24}{13}}{1 - \frac{144}{169}} = \dfrac{24}{13} \cdot \dfrac{169}{25} = \dfrac{312}{25} \).
In simple words: Divide through by cos² θ to get tan θ.

Teacher's Note:
1. 1 − 144/169 = 25/169.

Question 4. Prove that \( \sqrt{\dfrac{\operatorname{cosec}A - 1}{\operatorname{cosec}A + 1}} + \sqrt{\dfrac{\operatorname{cosec}A + 1}{\operatorname{cosec}A - 1}} = 2\sec A \). [CBSE 2015, 2016]
Answer:
Rationalise each root: \( \dfrac{\operatorname{cosec}A - 1}{\sqrt{\operatorname{cosec}^2A - 1}} + \dfrac{\operatorname{cosec}A + 1}{\sqrt{\operatorname{cosec}^2A - 1}} = \dfrac{2\operatorname{cosec}A}{\cot A} = \dfrac{2}{\cos A} = 2\sec A \).
In simple words: Multiply inside each root by the numerator.

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

Question 5. Prove that \( \dfrac{1 - \cos A}{1 + \cos A} = (\cot A - \operatorname{cosec}A)^2 \). [CBSE 2015]
Answer:
RHS \( = \left(\dfrac{\cos A - 1}{\sin A}\right)^2 = \dfrac{(1 - \cos A)^2}{1 - \cos^2A} = \dfrac{(1 - \cos A)^2}{(1 - \cos A)(1 + \cos A)} = \dfrac{1 - \cos A}{1 + \cos A} = \) LHS.
In simple words: Write everything in sin and cos.

Teacher's Note:
1. 1 − cos² A = (1 − cos A)(1 + cos A).

Question 6. If \( \theta = 45^\circ \), the value of \( \operatorname{cosec}^2\theta \) is [CBSE 2016]
(A) \( \dfrac{1}{2} \)   (B) 1   (C) \( -\dfrac{1}{2} \)   (D) 2
Answer:
Ans. (D) 2
\( \operatorname{cosec}45^\circ = \sqrt{2} \), so \( \operatorname{cosec}^2 45^\circ = 2 \).
In simple words: Use the table value.

Teacher's Note:
1. cosec 45° = √2.

Question 7. \( \sin(60^\circ + \theta) - \cos(30^\circ - \theta) \) is equal to [CBSE 2016]
(A) \( 2\cos\theta \)   (B) \( 2\sin\theta \)   (C) 1   (D) 0
Answer:
Ans. (D) 0
\( \cos(30^\circ - \theta) = \sin(90^\circ - 30^\circ + \theta) = \sin(60^\circ + \theta) \). The difference is 0.
In simple words: Convert cos to sin of the complement.

Teacher's Note:
1. The two angles are complementary.

Question 8. The value of \( [(\sec\theta + \tan\theta)(1 - \sin\theta)] \) is equal to [CBSE 2016]
(A) \( \tan^2\theta \)   (B) \( \sin^2\theta \)   (C) \( \cos\theta \)   (D) \( \sin\theta \)
Answer:
Ans. (C) \( \cos\theta \)
\( \left(\dfrac{1 + \sin\theta}{\cos\theta}\right)(1 - \sin\theta) = \dfrac{1 - \sin^2\theta}{\cos\theta} = \dfrac{\cos^2\theta}{\cos\theta} = \cos\theta \).
In simple words: Write sec and tan with cos.

Teacher's Note:
1. (1 + sin)(1 − sin) = cos².

Question 9. If A = 45° and B = 30°, then the value of \( \sin A\cos B + \cos A\sin B \) is [CBSE 2016]
(A) \( \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \)   (B) \( \dfrac{\sqrt{3} + 1}{2\sqrt{3}} \)   (C) \( \dfrac{\sqrt{3} - 1}{2\sqrt{2}} \)   (D) \( \dfrac{\sqrt{3} - 1}{2\sqrt{3}} \)
Answer:
Ans. (A) \( \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \)
\( \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{2} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \).
In simple words: Substitute the table values.

Teacher's Note:
1. Add the two products.

Question 10. If \( \cot\theta = \dfrac{7}{8} \), find the value of \( \dfrac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)} \). [CBSE 2016] Options: (A) 49/64 (B) 7/8 (C) 64/49 (D) 8/7
(A) 49/64   (B) 7/8   (C) 64/49   (D) 8/7
Answer:
Ans. (A) 49/64
\( \dfrac{1 - \sin^2\theta}{1 - \cos^2\theta} = \dfrac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta = \dfrac{49}{64} \).
In simple words: Use (1 + a)(1 − a) = 1 − a².

Teacher's Note:
1. The ratio is cot² θ.

Question 11. Prove that \( \sqrt{\dfrac{\operatorname{cosec}A - 1}{\operatorname{cosec}A + 1}} + \sqrt{\dfrac{\operatorname{cosec}A + 1}{\operatorname{cosec}A - 1}} = 2\sec A \). OR If ABC is a right angle triangle, right-angled at C, \( \angle A = 30^\circ \) and AB = 50 units, find the remaining two sides and \( \angle B \). [3 Marks, CBSE 2016]
Answer:
The first part is proved in the previous question.
OR: \( \angle B = 90^\circ - 30^\circ = 60^\circ \). \( BC = AB\sin 30^\circ = 25 \) units and \( AC = AB\cos 30^\circ = 25\sqrt{3} \) units.
\( \angle B = 60^\circ \), BC = 25, AC = \( 25\sqrt{3} \).
In simple words: Use sin 30° and cos 30° for the sides.

Teacher's Note:
1. BC is opposite the 30° angle, so it is half the hypotenuse.

Question 12. Prove that \( \dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \tan\theta + \cot\theta \). [CBSE 2016, 2017]
Answer:
\( \dfrac{\tan\theta}{1 - \cot\theta} = \dfrac{\tan^2\theta}{\tan\theta - 1} \); also \( \dfrac{\cot\theta}{1 - \tan\theta} = \dfrac{-1}{\tan\theta(\tan\theta - 1)} \).
Sum \( = \dfrac{\tan^3\theta - 1}{\tan\theta(\tan\theta - 1)} = \dfrac{(\tan\theta - 1)(\tan^2\theta + \tan\theta + 1)}{\tan\theta(\tan\theta - 1)} = \tan\theta + 1 + \cot\theta \). Hence proved.
In simple words: Write everything in tan θ.

Teacher's Note:
1. Use a³ − 1 = (a − 1)(a² + a + 1).

Question 13. Prove that \( \dfrac{1 - \sin\theta}{1 + \sin\theta} = (\sec\theta - \tan\theta)^2 \). [CBSE 2016, 2017]
Answer:
RHS \( = \left(\dfrac{1 - \sin\theta}{\cos\theta}\right)^2 = \dfrac{(1 - \sin\theta)^2}{1 - \sin^2\theta} = \dfrac{(1 - \sin\theta)^2}{(1 - \sin\theta)(1 + \sin\theta)} = \dfrac{1 - \sin\theta}{1 + \sin\theta} = \) LHS.
In simple words: Write in sin and cos.

Teacher's Note:
1. 1 − sin² = cos².

Question 14. Prove that \( \tan^2A - \tan^2B = \dfrac{\cos^2B - \cos^2A}{\cos^2B\cos^2A} = \dfrac{\sin^2A - \sin^2B}{\cos^2A\cos^2B} \). [CBSE 2016]
Answer:
\( \tan^2A - \tan^2B = \dfrac{\sin^2A}{\cos^2A} - \dfrac{\sin^2B}{\cos^2B} = \dfrac{\sin^2A\cos^2B - \sin^2B\cos^2A}{\cos^2A\cos^2B} \).
Numerator \( = \sin^2A(1 - \sin^2B) - \sin^2B(1 - \sin^2A) = \sin^2A - \sin^2B \). So the last expression holds.
Also \( \sin^2A - \sin^2B = (1 - \cos^2A) - (1 - \cos^2B) = \cos^2B - \cos^2A \), which gives the middle expression. Hence proved.
In simple words: Combine the fractions; cos² B cancels in the product.

Teacher's Note:
1. sin² = 1 − cos².

Question 15. Prove that \( (\operatorname{cosec}A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A} \). [2 Marks, CBSE 2016]
Answer:
LHS \( = \dfrac{\cos^2A}{\sin A} \cdot \dfrac{\sin^2A}{\cos A} = \sin A\cos A \). RHS \( = \dfrac{1}{\frac{\sin^2A + \cos^2A}{\sin A\cos A}} = \sin A\cos A \). So LHS = RHS.
In simple words: Both sides equal sin A cos A.

Teacher's Note:
1. tan A + cot A = 1/(sin A cos A).

Question 16. If \( \cos\theta + \cos^2\theta = 1 \), the value of \( \sin^2\theta + \sin^4\theta \) is [1 Mark, CBSE 2017]
(A) 0   (B) 1   (C) 2   (D) −1
Answer:
Ans. (B) 1
\( \cos\theta = 1 - \cos^2\theta = \sin^2\theta \). So \( \sin^2\theta + \sin^4\theta = \cos\theta + \cos^2\theta = 1 \).
In simple words: The given relation says cos θ = sin² θ.

Teacher's Note:
1. Replace sin² θ by cos θ.

Question 17. If \( \sec x + \tan x = p \), then \( \sec x \) is equal to [CBSE 2017]
(A) \( \dfrac{p^2 - 1}{p} \)   (B) \( \dfrac{p^2 + 1}{p} \)   (C) \( \dfrac{p^2 - 1}{2p} \)   (D) \( \dfrac{p^2 + 1}{2p} \)
Answer:
Ans. (D) \( \dfrac{p^2 + 1}{2p} \)
\( \sec x - \tan x = \dfrac{1}{p} \). Adding: \( 2\sec x = p + \dfrac{1}{p} \Rightarrow \sec x = \dfrac{p^2 + 1}{2p} \).
In simple words: Find sec x − tan x from the identity.

Teacher's Note:
1. Add the two equations.

Question 18. If \( \cos x = \cos 60^\circ\cos 30^\circ + \sin 60^\circ\sin 30^\circ \), then the value of x is [1 Mark, CBSE 2017]
(A) 90°   (B) 45°   (C) 30°   (D) 60°
Answer:
Ans. (C) 30°
\( \cos 60^\circ\cos 30^\circ + \sin 60^\circ\sin 30^\circ = \dfrac{1}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{3}}{2} = \cos 30^\circ \). So x = 30°.
In simple words: Substitute the table values.

Teacher's Note:
1. The two products are equal.

Question 19. Prove that \( \dfrac{\cos\theta}{1 + \sin\theta} = \dfrac{1 - \sin\theta}{\cos\theta} \). [3 Marks, CBSE 2017]
Answer:
Cross-multiplying: \( \cos^2\theta = (1 - \sin\theta)(1 + \sin\theta) = 1 - \sin^2\theta \), which is true. Hence proved.
In simple words: Cross-multiply.

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

Question 20. OR: Prove that (i) \( \tan 20^\circ\tan 35^\circ\tan 45^\circ\tan 55^\circ\tan 70^\circ = 1 \); (ii) \( \sin 48^\circ\sec 42^\circ + \operatorname{cosec}42^\circ\cos 48^\circ = 2 \); (iii) \( \dfrac{\sin 70^\circ}{\cos 20^\circ} + \dfrac{\operatorname{cosec}20^\circ}{\sec 70^\circ} - 2\cos 70^\circ\operatorname{cosec}20^\circ = 0 \). [3 Marks, CBSE 2017]
Answer:
(i) \( \tan 70^\circ = \cot 20^\circ \) and \( \tan 55^\circ = \cot 35^\circ \), so the product is \( 1 \times 1 \times \tan 45^\circ = 1 \).
(ii) \( \sec 42^\circ = \operatorname{cosec}48^\circ \), so \( \sin 48^\circ\operatorname{cosec}48^\circ + \operatorname{cosec}42^\circ\sin 42^\circ = 1 + 1 = 2 \).
(iii) \( \dfrac{\sin 70^\circ}{\cos 20^\circ} = 1 \), \( \dfrac{\operatorname{cosec}20^\circ}{\sec 70^\circ} = 1 \) and \( \cos 70^\circ\operatorname{cosec}20^\circ = 1 \). The value is \( 1 + 1 - 2 = 0 \).
In simple words: Use complementary angles.

Teacher's Note:
1. Each pair of ratios reduces to 1.

Question 21. Prove that \( \dfrac{\tan 45^\circ}{\operatorname{cosec}30^\circ} + \dfrac{\sec 60^\circ}{\cot 45^\circ} = \dfrac{5\sin 90^\circ}{2\cos 0^\circ} \). [2 Marks, CBSE 2017]
Answer:
LHS \( = \dfrac{1}{2} + \dfrac{2}{1} = \dfrac{5}{2} \). RHS \( = \dfrac{5 \times 1}{2 \times 1} = \dfrac{5}{2} \). So LHS = RHS.
In simple words: Substitute the table values on both sides.

Teacher's Note:
1. cosec 30° = 2 and sec 60° = 2.