CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set 01

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Question. (sin 30° + cos 30°) – (sin 60° + cos 60°)
(a) – 1
(b) 0
(c) 1
(d) 2
Answer: (b)
Explanation: According to question
\[ (\sin 30^\circ + \cos 30^\circ) - (\sin 60^\circ + \cos 60^\circ) = (\frac{1}{2} + \frac{\sqrt{3}}{2}) - (\frac{\sqrt{3}}{2} + \frac{1}{2}) = 0 \]

Question. Value of tan 30°/cot 60° is:
(a) 1/\( \sqrt{2} \)
(b) 1/\( \sqrt{3} \)
(c) \( \sqrt{3} \)
(d) 1
Answer: (d)
Explanation: \[ \frac{\tan 30^\circ}{\cot 60^\circ} = \frac{1/\sqrt{3}}{1/\sqrt{3}} = 1 \]

Question. sec²θ – 1 = ?
(a) tan²θ
(b) tan²θ + 1
(c) cot²θ – 1
(d) cos²θ
Answer: (a)
Explanation: From trigonometric identity \( 1+ \tan^2\theta = \sec^2\theta \Rightarrow \sec^2\theta – 1 = \tan^2\theta \)

Question. The value of sin θ and cos (90° – θ)
(a) Are same
(b) Are different
(c) No relation
(d) Information insufficient
Answer: (a)
Explanation: Since from trigonometric identities, \( \cos(90^\circ – \theta) = \sin \theta \). So, both represents the same value.

Question. If cos A = 4/5, then tan A = ?
(a) 3/5
(b) 3/4
(c) 4/3
(d) 4/5
Answer: (b)
Explanation: From trigonometric identity \( 1 + \tan^2 A = \sec^2 A \Rightarrow \sec^2 A – 1 = \tan^2 A \)
\[ \Rightarrow (\frac{5}{4})^2 - 1 = \tan^2 A \Rightarrow \frac{9}{16} = \tan^2 A \Rightarrow \tan A = \frac{3}{4} \]

Question. The value of the expression [cosec (75° + θ) – sec (15° - θ) – tan (55° + θ) + cot (35° - θ)] is
(a) 1
(b) –1
(c) 0
(d) 1/2
Answer: (c)
Explanation: Since
\( \csc (75^\circ + \theta) – \sec (15^\circ - \theta) – \tan (55^\circ + \theta) + \cot (35^\circ - \theta) \)
\( = \csc (75^\circ + \theta) – \csc [90^\circ - (15^\circ - \theta)] – \tan (55^\circ + \theta) + \tan [90^\circ - (35^\circ - \theta)] \)
\( = \csc (75^\circ + \theta) – \csc (75^\circ + \theta) – \tan (55^\circ + \theta) + \tan (55^\circ + \theta) = 0 \)

Question. Given that: SinA = a/b, then cosA = ?
(a) \( \frac{\sqrt{b^2+a^2}}{b} \)
(b) \( \frac{\sqrt{b^2-a^2}}{b} \)
(c) b/a
(d) a/b
Answer: (b)
Explanation: We have given: \( \sin A = \frac{a}{b} \). Let the complete ratio be x. Perpendicular = ax, Hypotenuse = bx. Base\(^2\) = Hypotenuse\(^2\) - Perpendicular\(^2\). Base = \( x\sqrt{b^2-a^2} \).
\[ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{x\sqrt{b^2-a^2}}{bx} = \frac{\sqrt{b^2-a^2}}{b} \]

Question. The value of (tan1° tan2° tan3° ... tan89°) is
(a) 0
(b) 1
(c) 2
(d) 1/2
Answer: (b)
Explanation: This can be written as, \( (\tan 1^\circ \tan 2^\circ \dots \tan 44^\circ \tan 45^\circ \tan 46^\circ \dots \tan 89^\circ) \)
\( = [\tan 1^\circ \tan 2^\circ \dots \tan 44^\circ \cdot 1 \cdot \cot 44^\circ \dots \cot 2^\circ \cot 1^\circ] = 1 \)

Question. If sin A + sin² A = 1, then cos² A + cos⁴ A = ?
(a) 1
(b) 0
(c) 2
(d) 4
Answer: (a)
Explanation: We have \( \sin A + \sin^2 A = 1 \Rightarrow \sin A = 1 – \sin^2 A \Rightarrow \sin A = \cos^2 A \). Squaring both sides: \( \sin^2 A = \cos^4 A \).
Therefore, \( \cos^2 A + \cos^4 A = \sin A + \sin^2 A = 1 \).

Question. If sin A = 1/2 and cos B = 1/2, then A + B = ?
(a) \( 0^\circ \)
(b) \( 30^\circ \)
(c) \( 60^\circ \)
(d) \( 90^\circ \)
Answer: (d)
Explanation: Since \( \sin A = 1/2 \Rightarrow \sin A = \sin 30^\circ \Rightarrow A = 30^\circ \).
And \( \cos B = 1/2 \Rightarrow \cos B = \cos 60^\circ \Rightarrow B = 60^\circ \).
Therefore \( A + B = 30^\circ + 60^\circ = 90^\circ \).

Question. The value of \( \frac{\sin^2 22^\circ + \sin^2 68^\circ}{\cos^2 22^\circ + \cos^2 68^\circ} + \sin^2 63^\circ + \cos 63^\circ \sin 27^\circ \) is:
(a) 3
(b) 2
(c) 1
(d) 0
Answer: (b)
Explanation: Using trigonometric properties, we have:
\[ \frac{\sin^2 22^\circ + \sin^2(90^\circ - 22^\circ)}{\cos^2(90^\circ - 68^\circ) + \cos^2 68^\circ} + \sin^2 63^\circ + \cos 63^\circ \sin(90^\circ - 63^\circ) \]
\[ = \frac{\sin^2 22^\circ + \cos^2 22^\circ}{\sin^2 68^\circ + \cos^2 68^\circ} + \sin^2 63^\circ + \cos^2 63^\circ = 1 + 1 = 2 \]

Question. If cos9α = sin α and 9α < 90°, then the value of tan 5α is
(a) \( \sqrt{3} \)
(b) 1/\( \sqrt{3} \)
(c) 0
(d) 1
Answer: (d)
Explanation: Since \( \cos 9\alpha = \sin \alpha \Rightarrow \sin (90^\circ - 9\alpha) = \sin \alpha \Rightarrow 90^\circ - 9\alpha = \alpha \Rightarrow \alpha = 9^\circ \).
Therefore, \( \tan 5\alpha = \tan 5 (9^\circ) = \tan 45^\circ = 1 \).

Question. If a pole 6m high casts a shadow \( 2\sqrt{3} \) m long on the ground, then the sun’s elevation is
(a) 60°
(b) 45°
(c) 30°
(d) 90°
Answer: (a)
Explanation: Let elevation be \( \theta \). Then \( \tan \theta = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \). Thus \( \theta = 60^\circ \).

Question. If cos (A + B) = 0, then sin (A – B) is reduced to:
(a) cos A
(b) cos 2B
(c) sin A
(d) sin 2B
Answer: (b)
Explanation: Since \( \cos (A + B) = 0 \Rightarrow A + B = 90^\circ \Rightarrow A = 90^\circ - B \).
This implies \( \sin (A – B) = \sin (90^\circ - B - B) = \sin (90^\circ - 2B) = \cos 2B \).

Question. If 4 tan A = 3, then \( \frac{4 \sin A - \cos A}{4 \sin A + \cos A} = ? \)
(a) 2/3
(b) 1/3
(c) 1/2
(d) 3/4
Answer: (c)
Explanation: Dividing numerator and denominator by \( \cos A \):
\[ \frac{4 \tan A - 1}{4 \tan A + 1} = \frac{3-1}{3+1} = \frac{2}{4} = \frac{1}{2} \]

Question. If \( x = \sin^2 \theta \) and \( y = \cos^2 \theta + 1 \), then find the value of x+y.
Answer: 2

Question. If \( \sec^2 \theta(1 + \sin \theta)(1 - \sin \theta) = k \), then find the value of k.
Answer: 1

Question. Find the value of \( \sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ \)
Answer: 0

Question. If \( \tan \theta \tan 45^\circ = 1 \), then find the value of \( \theta \)
Answer: \( 45^\circ \)

Question. Find the value of \( \sin^2 10^\circ + \sin^2 80^\circ \)
Answer: 1

Question. If \( \cos A = \frac{3}{5} \), then find the value of \( \tan^2 A - \sec^2 A \).
Answer: -1

Question. If \( \sin 2A = \cos 3A \), then find the value of A.
Answer: 18

Question. If \( \tan \theta = \tan(90 - \theta) \), then find the value of \( \theta \)
Answer: \( 45^\circ \)

Question. Complete the following:-
The angle nearer to altitude is ____________ than the angle away from the altitude.
Answer: greater

Question. If \( x = 3\sec^2 \theta - 1 \) and \( y = 3\tan^2 \theta - 2 \) then find the value of x – y.
Answer: 4

Question. If \( 2x = \csc \theta \) and \( \frac{2}{x} = \cot \theta \), then find the value of \( 4\left(x^2 - \frac{1}{x^2}\right) \)
Answer: 1

Question. If \( \sin \theta = \frac{3}{5} \), then find the value of \( 5\cos \theta \times \sin \theta \)
Answer: \( \frac{12}{5} \)

Question. If \( \tan \theta = \frac{12}{5} \) then find the value of \( \frac{13\sin \theta}{3} \)
Answer: 4

Question. Find the value of \( \frac{\csc 39^\circ}{\sec 51^\circ} + 2(\sin^2 5^\circ + \sin^2 85^\circ) \)
Answer: 3

Question. Find the value of \( \cos^2 15^\circ + \cos^2 25^\circ + \cos^2 65^\circ + \cos^2 75^\circ \)
Answer: 2

Question. Find the value of \( \sin^2 10^\circ + \sin^2 80^\circ \)
Answer: 1

Question. Find the value of \( \cos^2 67^\circ - \sin^2 23^\circ \)
Answer: 0

Question. Find the value of \( \tan 10^\circ \tan 20^\circ \tan 70^\circ \tan 80^\circ \)
Answer: 1

Question. Find the value of \( \csc A \sec(90^\circ - A) - \cot A \tan(90^\circ - A) \)
Answer: 1

Question. Find the value of \( \frac{\sin \theta - \sin^3 \theta}{\cos \theta - \cos^3 \theta} \)
Answer: \( \cot \theta \)

Question. If \( \tan \alpha = \frac{1}{\sqrt{3}} \) and \( \sin \beta = \frac{1}{\sqrt{2}} \), find the value of \( \alpha + \beta \).
Answer: \( 75^\circ \)

Question. If \( \csc \theta = 2 \) and \( \cot \theta = \sqrt{3}k \), then find the value of k.
Answer: 1

Question. If \( \csc^2 \theta = \frac{3}{2} \), then find the value of \( 2(\csc^2 \theta + \cot^2 \theta) \).
Answer: 4

Question. If \( \tan \theta = 4 \), then find the value of \( \frac{1}{10}(\tan^2 \theta + 2\sec^2 \theta) \)
Answer: 5

Question. If \( \sin \theta = \frac{1}{3} \), then find the value of \( 2\csc^2 \theta + \cot^2 \theta + 1 \).
Answer: 27

Question. If \( \cos \theta = \frac{\sqrt{3}}{2} \), then find the value of \( 8\sec^2 \theta + \tan^2 \theta + 1 \).
Answer: 12

Question. If \( 1 + 2\sin^2 \theta \cos^2 \theta = \sin^2 \theta + \cos^2 \theta + 4k \sin^2 \theta \cos^2 \theta \), then find the value of k.
Answer: \( \frac{1}{2} \)

Question. If \( \frac{\cos^2 20^\circ + \cos^2 70^\circ}{2(\sin^2 59^\circ + \sin^2 31^\circ)} = \frac{2}{k} \), then find the value of k.
Answer: 4

Question. Find the value of \( \tan 5^\circ \times \tan 30^\circ \times 4 \tan 85^\circ \)
Answer: \( \frac{4}{\sqrt{3}} \)

Question. If \( \frac{\cos 20^\circ}{\sin 70^\circ} + \frac{2\cos \theta}{\sin(90^\circ - \theta)} = \frac{k}{2} \), then find the value of k.
Answer: 6

Question. If \( \tan 4\theta = \cot \theta \), where \( 4\theta \) and \( \theta \) and are acute angles, then find the value of \( \theta \).
Answer: \( 18^\circ \)

Question. If \( \cos(81^\circ + \theta) = \sin\left(\frac{k}{3} - \theta\right) \), then find the value of k.
Answer: 27

Question. If \( \sec A = \frac{3}{2} \), then find the value of \( \tan^2 A \).
Answer: \( \frac{5}{4} \)

Question. If \( \cos 3\theta = 1 \), then find the value of \( \theta \).
Answer: \( 0^\circ \)

Question. If, A, B and C are the angles of a triangle, then find the value of \( \tan\left(\frac{A + B}{2}\right) \) in terms of angle C.
Answer: \( \cot \frac{C}{2} \)

Question. If \( \sin 3\theta = \cos 4\theta \), then find the value of \( 7\theta \).
Answer: \( 90^\circ \)

Question. At a point 30m. away form the foot of a tower the angle of elevation of the top of the tower is \( 60^\circ \). Find the height of the tower.
Answer: \( 30\sqrt{3} \)

Question. A person standing on the bank of a river observe that the angle of elevation of tree is \( 60^\circ \). When he moves 40m away, the angle of elevation becomes \( 30^\circ \). At what distance is he now standing away from tree?
Answer: 60m

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