CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 11

Question. If tan A = \( \frac{4}{3} \), the value of sin A is
(a) \( \frac{4}{5} \)
(b) \( \frac{3}{4} \)
(c) \( \frac{5}{3} \)
(d) \( \frac{7}{5} \)
Answer: (a) \( \frac{4}{5} \)
Explanation: tan A = \( \frac{4}{3} \)
\( \implies \) \( \tan^2 A = \frac{16}{9} \)
\( \implies \) \( \cot^2 A = \frac{9}{16} \)
\( \therefore \) \( \text{cosec}^2 A - 1 = \frac{9}{16} \)
\( \implies \) \( \text{cosec}^2 A = \frac{25}{16} \)
\( \implies \) \( \text{cosec } A = \frac{5}{4} \)
\( \implies \) \( \sin A = \frac{4}{5} \)

Question. If 3 tan A = 4, then the value of \( \frac{3 \sin A + 2 \cos A}{3 \sin A - 2 \cos A} \) is:
(a) 4
(b) \( \frac{11}{15} \)
(c) \( \frac{7}{15} \)
(d) 3
Answer: (d) 3
Explanation: 3 tan A = 4 \( \implies \) tan A = \( \frac{4}{3} \)
\( \therefore \) \( \frac{3 \sin A + 2 \cos A}{3 \sin A - 2 \cos A} \)
Divide numerator and denominator by cos A
\( \implies \) \( \frac{\frac{3 \sin A}{\cos A} + \frac{2 \cos A}{\cos A}}{\frac{3 \sin A}{\cos A} - \frac{2 \cos A}{\cos A}} = \frac{3 \tan A + 2}{3 \tan A - 2} \)
\( \implies \) \( \frac{3 \times \frac{4}{3} + 2}{3 \times \frac{4}{3} - 2} = \frac{4 + 2}{4 - 2} = \frac{6}{2} = 3 \)

Question. If sin \( \theta \) + cos \( \theta \) = \( \sqrt{2} \) cos \( \theta \), (\( \theta \neq 90^\circ \)) then the value of tan \( \theta \) is:
(a) \( \sqrt{2} - 1 \)
(b) \( \sqrt{2} + 1 \)
(c) \( \sqrt{2} \)
(d) \( -\sqrt{2} \)
Answer: (a) \( \sqrt{2} - 1 \)

Question. Given that sin \( \alpha = \frac{\sqrt{3}}{2} \) and cos \( \beta = 0 \), then the value of \( \beta - \alpha \) is:
(a) \( 0^\circ \)
(b) \( 90^\circ \)
(c) \( 60^\circ \)
(d) \( 30^\circ \)
Answer: (d) \( 30^\circ \)
 

Question. If sin (A + B) = cos (A - B) = 1, then
(a) A = B = 0
(b) A = B = \( 45^\circ \)
(c) A = \( 60^\circ \), B = \( 30^\circ \)
(d) None of the options
Answer: (b) A = B = 45º
Explanation: sin (A + B) = cos (A - B) = 1
\( \implies \) sin (A + B) = 1, \( \implies \) sin (A + B) = sin 90º
\( \implies \) A + B = 90º ...(i)
\( \implies \) cos (A - B) = 1, \( \implies \) cos (A - B) = cos 0º
\( \implies \) A - B = 0 or A = B
Putting in (i), 2A = 90º \( \implies \) A = 45º
A = B = 45º

Question. If cos A = \( \frac{5}{13} \), find the value of tan A + cot A
(a) \( \frac{169}{60} \)
(b) \( \frac{12}{13} \)
(c) 1
(d) \( \frac{60}{169} \)
Answer: (a) \( \frac{169}{60} \)
 Explanation: cos A = \( \frac{5}{13} \)
\( \implies \) \( \cos^2 A = \frac{25}{169} \), 1 - \( \sin^2 A = \frac{25}{169} \)
\( \implies \) \( \sin^2 A = \frac{144}{169} \) or \( \sin A = \frac{12}{13} \)
\( \implies \) tan A = \( \frac{\sin A}{\cos A} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5} \)
\( \implies \) cot A = \( \frac{1}{\tan A} = \frac{5}{12} \)
tan A + cot A = \( \frac{12}{5} + \frac{5}{12} = \frac{169}{60} \)

Question. If 5x = sec \( \theta \) and \( \frac{5}{x} \) = tan \( \theta \) then find the value of \( 5(x^2 - \frac{1}{x^2}) \)
(a) 5
(b) \( \frac{1}{5} \)
(c) \( \frac{2}{5} \)
(d) 0
Answer: (b) \( \frac{1}{5} \)
Explanation: 5x = sec \( \theta \), \( \frac{5}{x} = \tan \theta \)
Squaring, \( 25x^2 = \sec^2 \theta \), \( \frac{25}{x^2} = \tan^2 \theta \)
Subtracting,
\( \implies \) \( 25x^2 - \frac{25}{x^2} = \sec^2 \theta - \tan^2 \theta \)
\( \implies \) \( 25(x^2 - \frac{1}{x^2}) = 1 \)
\( \implies \) \( 5(x^2 - \frac{1}{x^2}) = \frac{1}{5} \)

Question. If sin A + \( \sin^2 A \) = 1, then the value of the expression (\( \cos^2 A + \cos^4 A \)) is: 
(a) 1
(b) \( \frac{1}{2} \)
(c) 2
(d) 3
Answer: (a) 1
Explanation: We know that \( \sin^2 \theta + \cos^2 \theta = 1 \) ....(i)
Given: sin A + \( \sin^2 A \) = 1
\( \implies \) sin A = 1 - \( \sin^2 A \)
\( \implies \) sin A = \( \cos^2 A \) [\( \because \sin^2 \theta + \cos^2 \theta = 1 \)]
Squaring both sides
\( \implies \) \( \sin^2 A = \cos^4 A \)
\( \implies \) 1 - \( \cos^2 A = \cos^4 A \) [using (i)]
\( \implies \) \( \cos^2 A + \cos^4 A = 1 \)

Question. The value of (1 + cos \( \theta \)) (1 - cos \( \theta \)) \( \text{cosec}^2 \theta \) =
(a) 0
(b) 1
(c) \( \cos^2 \theta \)
(d) \( \sin^2 \theta \)
Answer: (b) 1
Explanation: (1 + cos \( \theta \)) (1 - cos \( \theta \)) \( \text{cosec}^2 \theta \)
= (1 - \( \cos^2 \theta \)) \( \text{cosec}^2 \theta \)
= \( \sin^2 \theta \text{ cosec}^2 \theta = 1 \)

Question. If sin \( \theta \) – cos \( \theta \) = 0, then the value of (\( \sin^4 \theta + \cos^4 \theta \)) is: 
(a) 1
(b) \( \frac{3}{4} \)
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{4} \)
Answer: (c) \( \frac{1}{2} \)
Explanation: We know that tan \( \theta = \frac{\sin \theta}{\cos \theta} \) and sin 45º = cos 45º = \( \frac{1}{\sqrt{2}} \)
Given: (sin \( \theta \) – cos \( \theta \)) = 0
\( \implies \) sin \( \theta \) = cos \( \theta \) \( \implies \) \( \frac{\sin \theta}{\cos \theta} = 1 \)
\( \implies \) tan \( \theta \) = 1
\( \implies \) tan \( \theta \) = tan 45º [\( \because \text{tan 45º = 1} \)]
\( \implies \) \( \theta = 45^\circ \)
Now, \( \sin^4 \theta + \cos^4 \theta = \sin^4 45^\circ + \cos^4 45^\circ \)
= \( (\frac{1}{\sqrt{2}})^4 + (\frac{1}{\sqrt{2}})^4 \) [\( \because \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \)]
= \( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \)

Question. If \( \theta = 45^\circ \) then sec \( \theta \) cot \( \theta \) – cosec \( \theta \) tan \( \theta \) is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a) 0
Explanation: sec \( \theta \) cot \( \theta \) – cosec \( \theta \) tan \( \theta \)
= \( \frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta} \cdot \frac{\sin \theta}{\cos \theta} \)
= \( \frac{1}{\sin \theta} - \frac{1}{\cos \theta} = \frac{\cos \theta - \sin \theta}{\sin \theta \cos \theta} \)
{At \( \theta = 45^\circ \), sin \( \theta \) = cos \( \theta \)}
= \( \frac{0}{\sin \theta \cos \theta} = 0 \)

Question. If x = a sin \( \theta \) and y = a cos \( \theta \), then the value of \( x^2 + y^2 \) is
(a) a
(b) \( a^2 \)
(c) 1
(d) \( b^2 \)
Answer: (b) \( a^2 \)
Explanation: x = a sin \( \theta \), y = a cos \( \theta \)
Squaring, \( x^2 = a^2 \sin^2 \theta \), \( y^2 = a^2 \cos^2 \theta \)
Adding \( x^2 + y^2 = a^2 \sin^2 \theta + a^2 \cos^2 \theta \)
= \( a^2 (\sin^2 \theta + \cos^2 \theta) = a^2 \)
{ \( \sin^2 \theta + \cos^2 \theta = 1 \)}

Question. 4 \( \tan^2 A \) – 4 \( \sec^2 A \) is equal to:
(a) 2
(b) 3
(c) 4
(d) -4
Answer: (d) -4
Explanation: 4 \( \tan^2 A \) – 4 \( \sec^2 A \)
= – 4 (\( \sec^2 A - \tan^2 A \))
= – 4 × 1
= – 4

Question. If 3 cos \( \theta \) = 1, then cosec \( \theta \) is equal to:
(a) \( 2\sqrt{2} \)
(b) \( \frac{3}{2\sqrt{2}} \)
(c) \( \frac{2\sqrt{3}}{3} \)
(d) \( \frac{4}{3\sqrt{2}} \)
Answer: (b) \( \frac{3}{2\sqrt{2}} \)
Explanation: 3 cos \( \theta \) = 1 gives cos \( \theta = \frac{1}{3} \)
Here B = 1, H = 3
\( \therefore \) P = \( \sqrt{H^2 - B^2} = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2} \)
cosec \( \theta = \frac{H}{P} = \frac{3}{\sqrt{8}} \) i.e. \( \frac{3}{2\sqrt{2}} \)

Question. If cosec \( \theta \) – cot \( \theta = \frac{1}{3} \), then the value of cosec \( \theta \) + cot \( \theta \) is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
Explanation: We know that: \( \text{cosec}^2 \theta - \cot^2 \theta = 1 \)
i.e. (cosec \( \theta \) + cot \( \theta \))(cosec \( \theta \) – cot \( \theta \)) = 1
\( \implies \) cosec \( \theta \) + cot \( \theta = \frac{1}{\text{cosec } \theta - \cot \theta} = \frac{1}{\frac{1}{3}} = 3 \)

Question. If 2 sin 2\( \theta = \sqrt{3} \), then the value of \( \theta \) is:
(a) \( 90^\circ \)
(b) \( 30^\circ \)
(c) \( 60^\circ \)
(d) \( 45^\circ \)
Answer: (b) \( 30^\circ \)
Explanation: 2 sin 2\( \theta = \sqrt{3} \implies \sin 2\theta = \frac{\sqrt{3}}{2} \)
\( \implies \) sin 2\( \theta \) = sin 60º
\( \implies \) 2\( \theta \) = 60º \( \implies \) \( \theta \) = 30º

Question. If the height and length of the shadow of a man are the same, then the angle of elevation of the sun is: 
(a) \( 45^\circ \)
(b) \( 60^\circ \)
(c) \( 90^\circ \)
(d) \( 120^\circ \)
Answer: (a) \( 45^\circ \)
Explanation: Let AB be the height of a man and BC be the shadow of a man.
AB = BC [Given]
In right angled \( \Delta ABC \),
tan \( \theta = \frac{AB}{BC} \)
\( \implies \frac{AB}{BC} = \tan \theta \)
\( \implies \) tan \( \theta \) = 1
\( \implies \) \( \theta = 45^\circ \).

Write True False

Question. cos \( \theta = \frac{a^2 + b^2}{2ab} \), where a and b are two distinct numbers such that ab > 0.
Answer: False.
Explanation: We know that \( (a - b)^2 > 0 \) [As square of any number is positive]
\( \implies a^2 + b^2 - 2ab > 0 \)
\( \implies a^2 + b^2 > 2ab \)
\( \implies \frac{a^2 + b^2}{2ab} > 1 \)
But cos \( \theta = \frac{a^2 + b^2}{2ab} \)
\( \implies \text{cos } \theta > 1 \)
which is not possible, since, – 1 \( \le \) cos \( \theta \le \) 1.
Hence, cos \( \theta \neq \frac{a^2 + b^2}{2ab} \)

Question. The angle of elevation of the top of a tower is 30º. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.
Answer: False.

Question. If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains unchanged. 
Answer: True.

Fill in the Blanks

Question. Simplest form of \( \frac{1 + \tan^2 A}{1 + \cot^2 A} \) is .................... . 
Answer: (tan²A)
Answer: Explanation: \( \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \tan^2 A \)

Question. If tan A = 1, then 2 sin A cos A = .................... 
Answer: 1
Explanation: Since, tan A = 1, \( \therefore \) A = 45º
\( \therefore \) 2 sin A cos A = 2 × sin 45º × cos 45º = \( 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1 \)

Question. In the figure, the angles of depressions from the observing postitions O1 and O2, respectively of the object A are ...................., .................... . 
Answer: (30º, 45º)
Explanation: Angle of depression from O1 is \( \angle O_2O_1A = 90^\circ - \angle AO_1C = 90^\circ - 60^\circ = 30^\circ \). Angle of depressions from O2 is \( \angle XO_2A = 45^\circ \)

Question. The value of \( (\sin^2 \theta + \frac{1}{1 + \tan^2 \theta}) \)....................
Answer: 1
Explanation: \( \sin^2 \theta + \frac{1}{1 + \tan^2 \theta} = \sin^2 \theta + \frac{1}{\sec^2 \theta} = \sin^2 \theta + \cos^2 \theta = 1 \)

Question. Simplest form of (1 – \( \cos^2 A \)) (1 + \( \cot^2 A \)) is .................... . 
Answer: 1
Explanation: (1 – \( \cos^2 A \)) (1 + \( \cot^2 A \)) = \( \sin^2 A \cdot \text{cosec}^2 A = \sin^2 A \cdot \frac{1}{\sin^2 A} = 1 \)

Question. If 3 sec \( \theta \) – 5 = 0 then cot \( \theta \) = .................... .
Answer: 3/4
Explanation: 3 sec \( \theta \) – 5 = 0 \( \implies \) 3 sec \( \theta \) = 5 \( \implies \) sec \( \theta = \frac{5}{3} \)
\( \implies \) \( \sec^2 \theta = \frac{25}{9} \)
\( \implies \) 1 + \( \tan^2 \theta = \frac{25}{9} \)
\( \implies \) \( \tan^2 \theta = \frac{16}{9} \)
\( \implies \) tan \( \theta = \frac{4}{3} \)
\( \implies \) cot \( \theta = \frac{3}{4} \)

Question. If sin \( \theta \) – cos \( \theta \) = 0, 0 \( \le \theta \le \) 90º then the value of \( \theta \) is .................... .
Answer: 45º
Explanation: sin \( \theta \) – cos \( \theta \) = 0 \( \implies \) sin \( \theta \) = cos \( \theta \). At \( \theta \) = 45º, sin \( \theta = \frac{1}{\sqrt{2}} \) and cos \( \theta = \frac{1}{\sqrt{2}} \). So, \( \theta \) = 45º.

Question. cos 1º cos 2º cos 3º ..... cos 180º = .................... .
Answer: 0
Explanation: cos 1º cos 2º cos 3º ..... cos 180º. cos 90º = 0 \( \therefore \) 0 × cos 1º cos 2º .... cos 180º = 0.

Question. If tan \( \theta = \sqrt{3} \), then sec \( \theta \) = ................... .
Answer: 2
Explanation: tan \( \theta = \sqrt{3} \) gives \( \theta \) = 60º. So, sec \( \theta \) = sec 60º = 2.

Question. Maximum value of \( \frac{1}{\text{cosec } \theta} \) is .................... .
Answer: 1
Explanation: \( \frac{1}{\text{cosec } \theta} = \sin \theta \). As, sin \( \theta \le 1 \), \( \frac{1}{\text{cosec } \theta} \le 1 \). Thus, maximum value of \( \frac{1}{\text{cosec } \theta} \) is 1.

Question. If tan \( \theta \) + cot \( \theta \) = 2, then the value of \( \tan^2 \theta + \cot^2 \theta \) is .................... .
Answer: 2
Explanation: \( \tan^2 \theta + \cot^2 \theta = (\tan \theta + \cot \theta)^2 - 2 \tan \theta \cot \theta = (2)^2 - 2 \times 1 = 2 \)