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NCERT Solutions · Class 10 · Chapter 8: Introduction to Trigonometry
NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2
Exercise 8.2: Ratios of standard angles. Learn the table: \(\sin 30^\circ = \tfrac12\), \(\sin 45^\circ = \tfrac{1}{\sqrt2}\), \(\sin 60^\circ = \tfrac{\sqrt3}{2}\), \(\cos\) in reverse order, \(\tan 30^\circ = \tfrac{1}{\sqrt3}\), \(\tan 45^\circ = 1\), \(\tan 60^\circ = \sqrt3\). Substitute, then simplify surds.
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Exercise 8.2 questions and solutions
Exercise 8.2, Question 1
Evaluate.
(i) \(\sin 60^\circ\cos 30^\circ + \sin 30^\circ\cos 60^\circ\)
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\[\tfrac{\sqrt3}{2} \cdot \tfrac{\sqrt3}{2} + \tfrac12 \cdot \tfrac12 = \tfrac34 + \tfrac14\] Answer: \(1\)
(ii) \(2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ\)
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\(2(1) + \tfrac34 - \tfrac34\). Answer: \(2\)
(iii) \(\dfrac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}\)
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\[\dfrac{\frac{1}{\sqrt2}}{\frac{2}{\sqrt3} + 2} = \dfrac{\sqrt3}{\sqrt2(2 + 2\sqrt3)}\] Rationalise: \[\begin{aligned}\dfrac{\sqrt3(2\sqrt3 - 2)}{\sqrt2(12 - 4)} &= \dfrac{6 - 2\sqrt3}{8\sqrt2} \\ &= \dfrac{3\sqrt2 - \sqrt6}{8}\end{aligned}\] Answer: \(\dfrac{3\sqrt2 - \sqrt6}{8}\)
(iv) \(\dfrac{\sin 30^\circ + \tan 45^\circ - \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}\)
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Numerator \[\tfrac12 + 1 - \tfrac{2}{\sqrt3} = \tfrac32 - \tfrac{2}{\sqrt3}\]; denominator \[\tfrac{2}{\sqrt3} + \tfrac12 + 1 = \tfrac32 + \tfrac{2}{\sqrt3}\] Multiply top and bottom by \(2\sqrt3\): \(\dfrac{3\sqrt3 - 4}{3\sqrt3 + 4}\); rationalise: \[\dfrac{(3\sqrt3 - 4)^2}{27 - 16} = \dfrac{43 - 24\sqrt3}{11}\] Answer: \(\dfrac{43 - 24\sqrt3}{11}\)
(v) \(\dfrac{5\cos^2 60^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}\)
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Numerator \[\tfrac54 + \tfrac{16}{3} - 1 = \tfrac{67}{12}\]; denominator \(= 1\). Answer: \(\tfrac{67}{12}\)
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Exercise 8.2, Question 2
Choose the correct option.
(i) \(\dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ} =\) (A) \(\sin 60^\circ\) (B) \(\cos 60^\circ\) (C) \(\tan 60^\circ\) (D) \(\sin 30^\circ\)
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\[\dfrac{2/\sqrt3}{4/3} = \dfrac{\sqrt3}{2}\] Answer: (A) \(\sin 60^\circ\)
(ii) \(\dfrac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ} =\) (A) \(\tan 90^\circ\) (B) 1 (C) \(\sin 45^\circ\) (D) 0
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\(\dfrac{1 - 1}{1 + 1} = 0\). Answer: (D) \(0\)
(iii) \(\sin 2A = 2\sin A\) is true when A = (A) \(0^\circ\) (B) \(30^\circ\) (C) \(45^\circ\) (D) \(60^\circ\)
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At \(0^\circ\) both sides are 0; at the others \(\sin 2A \ne 2\sin A\) (e.g. \(30^\circ\): \(\tfrac{\sqrt3}{2} \ne 1\)). Answer: (A) \(0^\circ\)
(iv) \(\dfrac{2\tan 30^\circ}{1 - \tan^2 30^\circ} =\) (A) \(\cos 60^\circ\) (B) \(\sin 60^\circ\) (C) \(\tan 60^\circ\) (D) \(\sin 30^\circ\)
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\(\dfrac{2/\sqrt3}{2/3} = \sqrt3\). Answer: (C) \(\tan 60^\circ\)
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Exercise 8.2, Question 3
\(\tan(A + B) = \sqrt3\) and \(\tan(A - B) = \tfrac{1}{\sqrt3}\), with \(0^\circ < A + B \le 90^\circ\) and \(A > B\). Find A and B.
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\(A + B = 60^\circ\) and \(A - B = 30^\circ\). Adding: \(2A = 90^\circ\). Answer: \(A = 45^\circ,\ B = 15^\circ\)
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Exercise 8.2, Question 4
True or false? Justify.
(i) \(\sin(A + B) = \sin A + \sin B\)
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\(A = B = 30^\circ\): \(\sin 60^\circ = \tfrac{\sqrt3}{2}\) but \(\sin 30^\circ + \sin 30^\circ = 1\). Answer: False
(ii) The value of \(\sin\theta\) increases as \(\theta\) increases (for \(0^\circ \le \theta \le 90^\circ\)).
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\[\begin{gathered}\sin 0^\circ = 0, \\ \sin 30^\circ = \tfrac12, \\ \sin 45^\circ \approx 0.71, \\ \sin 60^\circ \approx 0.87, \\ \sin 90^\circ = 1\end{gathered}\] Answer: True
(iii) The value of \(\cos\theta\) increases as \(\theta\) increases.
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\(\cos 0^\circ = 1\) but \(\cos 90^\circ = 0\): it decreases. Answer: False
(iv) \(\sin\theta = \cos\theta\) for all values of \(\theta\).
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Only at \(\theta = 45^\circ\); e.g. \(\sin 30^\circ \ne \cos 30^\circ\). Answer: False
(v) \(\cot A\) is not defined for \(A = 0^\circ\).
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\[\begin{aligned}\cot 0^\circ &= \dfrac{\cos 0^\circ}{\sin 0^\circ} \\ &= \dfrac10\end{aligned}\], undefined. Answer: True
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← Exercise 8.1 All Chapter 8 exercises Exercise 8.3 → Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in . Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log .