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Advanced Study Material for Class 10 Mathematics Chapter 8 Introduction to Trigonometry
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Class 10 Mathematics Chapter 8 Introduction to Trigonometry Notes and Questions
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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
Please click the link below to download CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set A.
Free study material for Chapter 8 Introduction to Trigonometry
CBSE Class 10 Mathematics Chapter 8 Introduction to Trigonometry Study Material
Students can find all the important study material for Chapter 8 Introduction to Trigonometry on this page. This collection includes detailed notes, Mind Maps for quick revision, and Sure Shot Questions that will come in your CBSE exams. This material has been strictly prepared on the latest 2026 syllabus for Class 10 Mathematics. Our expert teachers always suggest you to use these tools daily to make your learning easier and faster.
Chapter 8 Introduction to Trigonometry Expert Notes & Solved Exam Questions
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