CBSE Class 10 Mathematics Trigonometry Assignment Set 32

Exercise-2 Objective Questions

Question 1. If \( \cot\theta = \dfrac{1}{\sqrt{3}} \), then the value of \( \dfrac{1 - \cos^2\theta}{2 - \sin^2\theta} \) is
(A) \( \dfrac{1}{5} \)   (B) \( \dfrac{2}{5} \)   (C) \( \dfrac{3}{5} \)   (D) None
Answer:
Ans. (C) \( \dfrac{3}{5} \)
\( \theta = 60^\circ \), \( \sin^2\theta = \dfrac{3}{4} \). \( \dfrac{\sin^2\theta}{2 - \sin^2\theta} = \dfrac{3/4}{5/4} = \dfrac{3}{5} \).
In simple words: cot 60° = 1/√3.

Teacher's Note:
1. 1 − cos²θ = sin²θ.

Question 2. Let \( \tan A = \dfrac{\sqrt{3}}{4 - \sqrt{3}} \) and \( \tan B = \dfrac{\sqrt{3}}{4 + \sqrt{3}} \), then \( \dfrac{\tan A - \tan B}{1 + \tan A\tan B} \) is equal to
(A) 0.385   (B) 0.375   (C) 0.575   (D) 0.465
Answer:
Ans. (B) 0.375
\( \tan A - \tan B = \sqrt{3}\cdot\dfrac{2\sqrt{3}}{13} = \dfrac{6}{13} \) and \( \tan A\tan B = \dfrac{3}{13} \).
\( \dfrac{6/13}{16/13} = \dfrac{6}{16} = 0.375 \).
In simple words: (4 − √3)(4 + √3) = 13.

Teacher's Note:
1. Do the numerator and the product separately.

Question 3. Triangle PQR is right angled at Q and has side lengths PQ = 14 and QR = 48. If M is the mid-point of PR and \( \angle MQP = \theta \), then \( \cos\theta \) is equal to
(A) \( \dfrac{7}{25} \)   (B) \( \dfrac{7}{50} \)   (C) \( \dfrac{7}{24} \)   (D) \( \dfrac{24}{25} \)
Answer:
Ans. (A) \( \dfrac{7}{25} \)
\( PR = \sqrt{14^2 + 48^2} = 50 \). The mid-point of the hypotenuse is equidistant from all vertices, so \( MQ = MP \) and \( \angle MQP = \angle MPQ = \angle P \).
\( \cos\theta = \cos P = \dfrac{PQ}{PR} = \dfrac{14}{50} = \dfrac{7}{25} \).
In simple words: MQ = MP gives equal base angles.

Teacher's Note:
1. The mid-point of the hypotenuse is the circumcentre.

Question 4. Given \( 8\tan\theta = 3\cos\theta \), the value of \( E = \sin\theta + \cos\theta + \tan\theta + \sec\theta + \operatorname{cosec}\theta + \cot\theta \) can be expressed as \( \dfrac{a + b\sqrt{c}}{d} \), where b, c and d are primes. Then the value of (a + b + c + d) is equal to
(A) 26   (B) 25   (C) 24   (D) 23
Answer:
Ans. (A) 26
\( 8\sin\theta = 3\cos^2\theta = 3 - 3\sin^2\theta \Rightarrow 3\sin^2\theta + 8\sin\theta - 3 = 0 \Rightarrow \sin\theta = \dfrac{1}{3} \), \( \cos\theta = \dfrac{2\sqrt{2}}{3} \).
\( E = \dfrac{1}{3} + \dfrac{2\sqrt{2}}{3} + \dfrac{\sqrt{2}}{4} + \dfrac{3\sqrt{2}}{4} + 3 + 2\sqrt{2} = \dfrac{10 + 11\sqrt{2}}{3} \). So a = 10, b = 11, c = 2, d = 3 and the sum is 26.
In simple words: Reject the negative root of the quadratic.

Teacher's Note:
1. Find sin θ from the quadratic first.

Question 5. The value of the expression \( \dfrac{\sin 30^\circ + \tan 45^\circ - \sec 60^\circ}{\operatorname{cosec} 30^\circ - \cot 45^\circ - \cos 60^\circ} = \)
(A) 0   (B) 1   (C) \( -1 \)   (D) \( 2 + \sqrt{3} \)
Answer:
Ans. (C) \( -1 \)
Numerator \( = \dfrac{1}{2} + 1 - 2 = -\dfrac{1}{2} \). Denominator \( = 2 - 1 - \dfrac{1}{2} = \dfrac{1}{2} \). The value is \( -1 \).
In simple words: sec 60° = 2 and cosec 30° = 2.

Teacher's Note:
1. Substitute the standard values.

Teacher's Note:
1. Find y from the two angle equations.

Question 7. Which one of the following quantities is not rational?
(A) \( \dfrac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} \)   (B) \( 4\cos^3 30^\circ - 3\cos 30^\circ \)   (C) \( 3\sin 30^\circ - 4\sin^3 30^\circ \)   (D) \( \dfrac{2\cot 30^\circ}{\cot^2 30^\circ - 1} \)
Answer:
Ans. (D) \( \dfrac{2\cot 30^\circ}{\cot^2 30^\circ - 1} \)
(A) \( = \cos 60^\circ = \dfrac{1}{2} \). (B) \( = \cos 90^\circ = 0 \). (C) \( = \sin 90^\circ = 1 \).
(D) \( = \dfrac{2\sqrt{3}}{3 - 1} = \sqrt{3} \), which is irrational.
In simple words: Only (D) gives √3.

Teacher's Note:
1. Use the triple angle formulas or substitute directly.

Question 8. A square DEAF is constructed inside a 30°-60°-90° triangle ABC with hypotenuse BC = 4, D on side BC, E on side AC and F on side AB. The length of the side of the square is
(A) \( 3 - \sqrt{3} \)   (B) \( 3 - \sqrt{2} \)   (C) 2   (D) 1.5
Answer:
Ans. (A) \( 3 - \sqrt{3} \)
AB = 2 and \( AC = 2\sqrt{3} \) (right angle at A). The square has a corner at A, so \( s = \dfrac{AB \cdot AC}{AB + AC} = \dfrac{4\sqrt{3}}{2 + 2\sqrt{3}} = \dfrac{2\sqrt{3}}{1 + \sqrt{3}} \).
\( s = \dfrac{2\sqrt{3}(\sqrt{3} - 1)}{2} = 3 - \sqrt{3} \).
In simple words: The small triangle cut off by the square is similar to ABC.

Teacher's Note:
1. Use the inscribed square formula ab/(a + b).

Question 9. Let \( S = \sin^2 30^\circ + \sin^2 45^\circ + \sin^2 60^\circ \) and \( P = \operatorname{cosec}^2 45^\circ \cdot \sec^2 30^\circ \cdot \sin^3 90^\circ \cdot \cos 60^\circ \), then the correct statement is
(A) S < P   (B) S = P   (C) S × P = 2   (D) S + P > 3
Answer:
Ans. (C) S × P = 2
\( S = \dfrac{1}{4} + \dfrac{1}{2} + \dfrac{3}{4} = \dfrac{3}{2} \). \( P = 2 \times \dfrac{4}{3} \times 1 \times \dfrac{1}{2} = \dfrac{4}{3} \).
\( S \times P = \dfrac{3}{2} \times \dfrac{4}{3} = 2 \). (Also S > P and \( S + P = \dfrac{17}{6} < 3 \).)
In simple words: Check each option against S = 3/2 and P = 4/3.

Teacher's Note:
1. Compute S and P separately.

Question 10. Suppose that \( \theta \) is a number and \( 0 < \theta < 90^\circ \). Consider the statements I. \( \sec\theta^\circ < 1 \), II. \( \sec\theta^\circ = 1 \), III. \( \sec\theta^\circ > 1 \) and select the correct choice.
(A) Only I is possible   (B) All are possible   (C) Only III is possible   (D) Only I is impossible
Answer:
Ans. (C) Only III is possible
For \( 0^\circ < \theta < 90^\circ \), \( 0 < \cos\theta < 1 \), so \( \sec\theta = \dfrac{1}{\cos\theta} > 1 \). Only III is possible.
In simple words: sec 0° = 1, but 0° is excluded.

Teacher's Note:
1. Secant is the reciprocal of a number between 0 and 1.

Teacher's Note:
1. Find α and β from the two triangles.

Question 12. A right triangle has angles which measure 30°, 60° and 90°. If the perimeter of this triangle is \( 15 + 5\sqrt{3} \), then the length of the hypotenuse of this triangle is
(A) 5   (B) 7.5   (C) 10   (D) 12.5
Answer:
Ans. (C) 10
The sides are k, \( k\sqrt{3} \), 2k, so \( k(3 + \sqrt{3}) = 5(3 + \sqrt{3}) \Rightarrow k = 5 \). The hypotenuse is \( 2k = 10 \).
In simple words: Divide the perimeter by 3 + √3.

Teacher's Note:
1. Use the 1 : √3 : 2 ratio.

Question 13. Let \( T_1 = \dfrac{\sin 45^\circ - \sin 30^\circ + \cot 90^\circ}{\cos 45^\circ + \cos 60^\circ} \) and \( T_2 = \dfrac{\sec 45^\circ - \tan 45^\circ}{\operatorname{cosec} 45^\circ + \cos 0^\circ + \cot 90^\circ} \), then
(A) \( T_1 + T_2 = 0 \)   (B) \( T_1 - T_2 = 0 \)   (C) \( T_1 = 2T_2 \)   (D) \( T_1 = -T_2 \)
Answer:
Ans. (B) \( T_1 - T_2 = 0 \)
\( T_1 = \dfrac{\frac{\sqrt{2}}{2} - \frac{1}{2}}{\frac{\sqrt{2}}{2} + \frac{1}{2}} = \dfrac{\sqrt{2} - 1}{\sqrt{2} + 1} \). \( T_2 = \dfrac{\sqrt{2} - 1}{\sqrt{2} + 1} \).
So \( T_1 = T_2 \), that is \( T_1 - T_2 = 0 \) (each equals \( 3 - 2\sqrt{2} \)).
In simple words: cot 90° = 0 and cos 0° = 1.

Teacher's Note:
1. Simplify both fractions.

Question 14. Which one of the following when simplified is not equal to one?
(A) \( \tan 18^\circ \tan 36^\circ \tan 54^\circ \tan 72^\circ \)   (B) \( \sin^2 19^\circ + \sin^2 71^\circ \)   (C) \( \dfrac{2\sin 62^\circ}{\cos 28^\circ} - \dfrac{\sec 42^\circ}{\operatorname{cosec} 48^\circ} \)   (D) None of these
Answer:
Ans. (D) None of these
(A) pairs \( \tan 18^\circ\tan 72^\circ = 1 \) and \( \tan 36^\circ\tan 54^\circ = 1 \). (B) \( \sin^2 19^\circ + \cos^2 19^\circ = 1 \). (C) \( 2 - 1 = 1 \).
All three equal 1, so the answer is None of these.
In simple words: sin 62° = cos 28° and cosec 48° = sec 42°.

Teacher's Note:
1. Use complementary angles in each.

Question 15. Evaluate \( \sec^2 10^\circ - \cot^2 80^\circ + \dfrac{\sin 15^\circ\cos 75^\circ + \cos 15^\circ\sin 75^\circ}{\cos\theta\sin(90^\circ - \theta) + \sin\theta\cos(90^\circ - \theta)} \)
(A) 0   (B) 1   (C) 2   (D) 3
Answer:
Ans. (C) 2
\( \cot 80^\circ = \tan 10^\circ \), so \( \sec^2 10^\circ - \tan^2 10^\circ = 1 \). The numerator is \( \sin(15^\circ + 75^\circ) = 1 \). The denominator is \( \cos^2\theta + \sin^2\theta = 1 \).
Total \( = 1 + 1 = 2 \).
In simple words: sin 15° cos 75° + cos 15° sin 75° = sin²75° + cos²75° = 1, which avoids the sum formula.

Teacher's Note:
1. Use complementary angle conversions.

Question 16. \( \dfrac{1 + \sec\theta}{\sec\theta} = \)
(A) \( \dfrac{\sin^2\theta}{1 - \cos\theta} \)   (B) \( \dfrac{\sin^2\theta}{1 - \sin\theta} \)   (C) \( \dfrac{\cos^2\theta}{1 - \sin\theta} \)   (D) \( 1 + \sin\theta \)
Answer:
Ans. (A) \( \dfrac{\sin^2\theta}{1 - \cos\theta} \)
\( \dfrac{1 + \sec\theta}{\sec\theta} = \cos\theta + 1 = \dfrac{(1 + \cos\theta)(1 - \cos\theta)}{1 - \cos\theta} = \dfrac{\sin^2\theta}{1 - \cos\theta} \).
In simple words: 1 − cos²θ = sin²θ.

Teacher's Note:
1. Multiply and divide by 1 − cos θ.

Question 17. If \( \theta \) is an acute angle and \( \tan\theta + \cot\theta = 2 \), then
(A) \( \tan\theta = \dfrac{1}{2} \)   (B) \( \tan\theta = 2 \)   (C) \( \tan^5\theta + \cot^5\theta = 32 \)   (D) \( \tan^7\theta + \cot^7\theta = 2 \)
Answer:
Ans. (D) \( \tan^7\theta + \cot^7\theta = 2 \)
\( t + \dfrac{1}{t} = 2 \Rightarrow (t - 1)^2 = 0 \Rightarrow \tan\theta = 1 \).
Then \( \tan^7\theta + \cot^7\theta = 1 + 1 = 2 \); the other options fail (\( \tan^5\theta + \cot^5\theta = 2 \), not 32).
In simple words: t + 1/t = 2 gives t = 1.

Teacher's Note:
1. The sum of a number and its reciprocal is 2 only for 1.

Question 18. The value of \( \cos^4\theta + \sin^4\theta + 2\cos^2\theta\sin^2\theta \) when \( \theta = 60^\circ \) is
(A) 1   (B) \( 2 \)   (C) \( \dfrac{1}{\sqrt{2}} \)   (D) \( 2\sqrt{2} \)
Answer:
Ans. (A) 1
The expression is \( (\cos^2\theta + \sin^2\theta)^2 = 1 \) for every \( \theta \).
In simple words: (a + b)² = a² + 2ab + b².

Teacher's Note:
1. Spot the perfect square.

Question 19. The value of the expression \( (\cos\theta - 1)(1 + \cos\theta)(1 + \cot^2\theta) \) is
(A) 0   (B) 1   (C) \( \sin^2\theta \)   (D) \( -1 \)
Answer:
Ans. (D) \( -1 \)
\( (\cos^2\theta - 1)\operatorname{cosec}^2\theta = -\sin^2\theta \cdot \dfrac{1}{\sin^2\theta} = -1 \).
In simple words: 1 + cot²θ = cosec²θ.

Teacher's Note:
1. (cos θ − 1)(cos θ + 1) = −sin²θ.

Question 20. If \( \dfrac{\sin x}{1 + \sec x} + \dfrac{\sin x}{\sec x - 1} = 2 \), where \( 0^\circ < x < 90^\circ \), then \( \operatorname{cosec} x \) has the value equal to
(A) 1   (B) 2   (C) \( \sqrt{2} \)   (D) \( \sqrt{3} \)
Answer:
Ans. (C) \( \sqrt{2} \)
The sum is \( \sin x\cdot\dfrac{2\sec x}{\sec^2 x - 1} = \dfrac{2\sin x\sec x}{\tan^2 x} = 2\cot x \). So \( \cot x = 1 \), \( x = 45^\circ \).
\( \operatorname{cosec} 45^\circ = \sqrt{2} \).
In simple words: sec²x − 1 = tan²x.

Teacher's Note:
1. Combine the two fractions.

Question 21. If \( \cos\theta + \sin\theta = \sqrt{2}\cos\theta \), then \( \cos\theta - \sin\theta \) is equal to
(A) \( \sqrt{2} - 1 \)   (B) \( \sqrt{2} \)   (C) \( \sqrt{2}\sin\theta \)   (D) \( \sqrt{2} + \sin\theta \)
Answer:
Ans. (C) \( \sqrt{2}\sin\theta \)
\( \sin\theta = (\sqrt{2} - 1)\cos\theta \Rightarrow \cos\theta = (\sqrt{2} + 1)\sin\theta \).
\( \cos\theta - \sin\theta = (\sqrt{2} + 1 - 1)\sin\theta = \sqrt{2}\sin\theta \).
In simple words: 1/(√2 − 1) = √2 + 1.

Teacher's Note:
1. Express cos θ through sin θ.

Question 22. If \( x\sin^3\theta + y\cos^3\theta = \sin\theta\cos\theta \) and \( x\sin\theta = y\cos\theta \), then
(A) \( x^3 + y^3 = 1 \)   (B) \( x^2 - y^2 = 1 \)   (C) \( x^2 + y^2 = 1 \)   (D) \( x^3 - y^3 = 1 \)
Answer:
Ans. (C) \( x^2 + y^2 = 1 \)
Let \( x\sin\theta = y\cos\theta = k \). Then \( x\sin^3\theta + y\cos^3\theta = k(\sin^2\theta + \cos^2\theta) = k \), so \( k = \sin\theta\cos\theta \).
\( x = \cos\theta \) and \( y = \sin\theta \), so \( x^2 + y^2 = 1 \).
In simple words: sin²θ + cos²θ = 1 collapses the expression.

Teacher's Note:
1. Put the common value k.

Question 23. Let \( x = r\cos\alpha\cdot\cos\beta \), \( y = r\cos\alpha\cdot\sin\beta \) and \( z = r\sin\alpha \), then \( (x^2 + y^2 + z^2) \) is
(A) independent of both \( \alpha \) and \( \beta \)   (B) independent of \( \alpha \) but dependent on \( \beta \)   (C) independent of \( \beta \) but dependent on \( \alpha \)   (D) dependent on both \( \alpha \) and \( \beta \)
Answer:
Ans. (A) independent of both \( \alpha \) and \( \beta \)
\( x^2 + y^2 = r^2\cos^2\alpha \), so \( x^2 + y^2 + z^2 = r^2(\cos^2\alpha + \sin^2\alpha) = r^2 \), independent of both.
In simple words: The result is r².

Teacher's Note:
1. Add x² and y² first.

Question 24. Given \( 2y\cos\theta = x\sin\theta \) and \( 2x\sec\theta - y\operatorname{cosec}\theta = 3 \), then the value of \( x^2 + 4y^2 \) is equal to
(A) 1   (B) 2   (C) 3   (D) 4
Answer:
Ans. (D) 4
From the first, \( x = 2y\cot\theta \). The second: \( \dfrac{4y}{\sin\theta} - \dfrac{y}{\sin\theta} = 3 \Rightarrow y = \sin\theta \), and \( x = 2\cos\theta \).
\( x^2 + 4y^2 = 4\cos^2\theta + 4\sin^2\theta = 4 \).
In simple words: Substitute x = 2y cot θ.

Teacher's Note:
1. Solve for x and y in terms of θ.