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NCERT Solutions · Class 10 · Chapter 8: Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.3

Exercise 8.3: Trigonometric identities. Use \(\sin^2 A + \cos^2 A = 1\), \(\sec^2 A - \tan^2 A = 1\), \(\csc^2 A - \cot^2 A = 1\). To prove an identity, start from the more complicated side, write everything in \(\sin\) and \(\cos\) if stuck, and simplify to the other side.

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Exercise 8.3 questions and solutions

Exercise 8.3, Question 1

Express \(\sin A\), \(\sec A\) and \(\tan A\) in terms of \(\cot A\).
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  1. \(\csc^2 A = 1 + \cot^2 A\), so \(\sin A = \dfrac{1}{\sqrt{1 + \cot^2 A}}\).
  2. \(\tan A = \dfrac{1}{\cot A}\); \[\begin{aligned}\sec^2 A &= 1 + \tan^2 A \\ &= \dfrac{\cot^2 A + 1}{\cot^2 A}\end{aligned}\], so \[\sec A = \dfrac{\sqrt{1 + \cot^2 A}}{\cot A}\]
Answer: \[\begin{gathered}\sin A = \dfrac{1}{\sqrt{1 + \cot^2 A}}, \\ \sec A = \dfrac{\sqrt{1 + \cot^2 A}}{\cot A}, \\ \tan A = \dfrac{1}{\cot A}\end{gathered}\]

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Exercise 8.3, Question 2

Write all the other trigonometric ratios of \(\angle A\) in terms of \(\sec A\).
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  1. \(\cos A = \dfrac{1}{\sec A}\); \(\tan A = \sqrt{\sec^2 A - 1}\) (acute A).
  2. \[\begin{aligned}\sin A &= \tan A\cos A \\ &= \dfrac{\sqrt{\sec^2 A - 1}}{\sec A}\end{aligned}\]; \(\cot A = \dfrac{1}{\sqrt{\sec^2 A - 1}}\); \[\csc A = \dfrac{\sec A}{\sqrt{\sec^2 A - 1}}\]
Answer: \[\begin{gathered}\cos A = \tfrac{1}{\sec A}, \\ \sin A = \tfrac{\sqrt{\sec^2 A - 1}}{\sec A}, \\ \tan A = \sqrt{\sec^2 A - 1}, \\ \cot A = \tfrac{1}{\sqrt{\sec^2 A - 1}}, \\ \csc A = \tfrac{\sec A}{\sqrt{\sec^2 A - 1}}\end{gathered}\]

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Exercise 8.3, Question 3

Choose the correct option.
(i) \(9\sec^2 A - 9\tan^2 A =\) (A) 1 (B) 9 (C) 8 (D) 0
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  1. \(9(\sec^2 A - \tan^2 A) = 9 \times 1\).
Answer: (B) \(9\)
(ii) \((1 + \tan\theta + \sec\theta)(1 + \cot\theta - \csc\theta) =\) (A) 0 (B) 1 (C) 2 (D) −1
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  1. In sin and cos: \[\dfrac{(\cos\theta + \sin\theta + 1)(\sin\theta + \cos\theta - 1)}{\sin\theta\cos\theta}\]
  2. Numerator \[= (\sin\theta + \cos\theta)^2 - 1 = 2\sin\theta\cos\theta\]
Answer: (C) \(2\)
(iii) \((\sec A + \tan A)(1 - \sin A) =\) (A) \(\sec A\) (B) \(\sin A\) (C) \(\csc A\) (D) \(\cos A\)
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  1. \[\begin{aligned}\dfrac{1 + \sin A}{\cos A}(1 - \sin A) &= \dfrac{1 - \sin^2 A}{\cos A} \\ &= \dfrac{\cos^2 A}{\cos A}\end{aligned}\]
Answer: (D) \(\cos A\)
(iv) \(\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =\) (A) \(\sec^2 A\) (B) −1 (C) \(\cot^2 A\) (D) \(\tan^2 A\)
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  1. \[\dfrac{\sec^2 A}{\csc^2 A} = \dfrac{\sin^2 A}{\cos^2 A}\]
Answer: (D) \(\tan^2 A\)

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Exercise 8.3, Question 4

Prove the identities (where the expressions are defined).
(i) \((\csc\theta - \cot\theta)^2 = \dfrac{1 - \cos\theta}{1 + \cos\theta}\)
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  1. LHS \[= \left(\dfrac{1 - \cos\theta}{\sin\theta}\right)^2 = \dfrac{(1 - \cos\theta)^2}{1 - \cos^2\theta}\]
  2. \[= \dfrac{(1 - \cos\theta)^2}{(1 - \cos\theta)(1 + \cos\theta)} = \dfrac{1 - \cos\theta}{1 + \cos\theta}\]
Answer: Proved.
(ii) \(\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2\sec A\)
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  1. Common denominator: \[\dfrac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A}\]
  2. Numerator \[= \cos^2 A + 1 + 2\sin A + \sin^2 A = 2(1 + \sin A)\], so LHS \(= \dfrac{2}{\cos A}\).
Answer: Proved.
(iii) \(\dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \sec\theta\csc\theta\)
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  1. Let \(t = \tan\theta\): LHS \[\begin{aligned}= \dfrac{t^2}{t - 1} - \dfrac{1}{t(t - 1)} &= \dfrac{t^3 - 1}{t(t - 1)} \\ &= \dfrac{t^2 + t + 1}{t}\end{aligned}\]
  2. \[\begin{aligned}= t + 1 + \dfrac1t &= 1 + \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} \\ &= 1 + \sec\theta\csc\theta\end{aligned}\]
Answer: Proved.
(iv) \(\dfrac{1 + \sec A}{\sec A} = \dfrac{\sin^2 A}{1 - \cos A}\)
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  1. LHS \(= \cos A + 1\).
  2. RHS \[= \dfrac{1 - \cos^2 A}{1 - \cos A} = 1 + \cos A\]
Answer: Proved.
(v) \(\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A\)
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  1. Divide top and bottom by \(\sin A\): \[\dfrac{\cot A + \csc A - 1}{\cot A - \csc A + 1}\]
  2. Replace the 1 on top by \(\csc^2 A - \cot^2 A\); the numerator factorises: \[(\csc A + \cot A)(1 - \csc A + \cot A)\]
  3. The bracket cancels with the denominator.
Answer: Proved.
(vi) \(\sqrt{\dfrac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A\)
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  1. Multiply inside by \(\dfrac{1 + \sin A}{1 + \sin A}\): \[\sqrt{\dfrac{(1 + \sin A)^2}{\cos^2 A}} = \dfrac{1 + \sin A}{\cos A}\] (positive for acute A).
Answer: Proved.
(vii) \(\dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta\)
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  1. \[\dfrac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(2\cos^2\theta - 1)}\], and \(1 - 2\sin^2\theta = 2\cos^2\theta - 1\).
  2. (Undefined where \(2\cos^2\theta = 1\), i.e. \(\theta = 45^\circ\).)
Answer: Proved.
(viii) \((\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A\)
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  1. Expand: \[\sin^2 A + 2 + \csc^2 A + \cos^2 A + 2 + \sec^2 A\]
  2. \[= 5 + (1 + \cot^2 A) + (1 + \tan^2 A) = 7 + \tan^2 A + \cot^2 A\]
Answer: Proved.
(ix) \((\csc A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}\)
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  1. LHS \[= \dfrac{\cos^2 A}{\sin A} \cdot \dfrac{\sin^2 A}{\cos A} = \sin A\cos A\]
  2. RHS \[= \dfrac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A\cos A}} = \sin A\cos A\]
Answer: Proved.
(x) \(\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \left(\dfrac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A\)
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  1. First: \(\dfrac{\sec^2 A}{\csc^2 A} = \tan^2 A\).
  2. Second: \[\begin{aligned}\dfrac{1 - \tan A}{1 - \frac{1}{\tan A}} &= \dfrac{(1 - \tan A)\tan A}{\tan A - 1} \\ &= -\tan A\end{aligned}\]; squared, \(\tan^2 A\).
Answer: Proved.

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