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If Sin − 1 ( 2 a 1 − a 2 ) + Cos − 1 ( 1 − a 2 1 + a 2 ) = Tan − 1 ( 2 X 1 − X 2 ) , Where a , X ∈ ( 0 , 1 ) , Then, the Value of X is (A) 0 (B) a 2 (C) a (D) 2 a 1 − a 2 - Mathematics

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Question

If \[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right),\text{ where }a, x \in \left( 0, 1 \right)\] , then, the value of x is

 

Options

  • 0

  • `a/2`

  •  a

  • `(2a)/(1-a^2)`

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Solution

\[\sin^{- 1} \left( \frac{2a}{1 - a^2} \right) + \cos^{- 1} \left( \frac{1 - a^2}{1 + a^2} \right) = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\]
\[ \Rightarrow 2 \tan^{- 1} a + 2 \tan^{- 1} a = 2 \tan^{- 1} x\]
\[ \Rightarrow 4 \tan^{- 1} a = 2 \tan^{- 1} x\]
\[ \Rightarrow 2 \tan^{- 1} a = \tan^{- 1} x\]
\[ \Rightarrow \tan^{- 1} \left( \frac{2a}{1 - a^2} \right) = \tan^{- 1} x\]
\[ \Rightarrow x = \frac{2a}{1 - a^2}\]

Hence, the correct answer is option(d).

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Chapter 4: Inverse Trigonometric Functions - Exercise 4.16 [Page 122]

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RD Sharma Mathematics [English] Class 12
Chapter 4 Inverse Trigonometric Functions
Exercise 4.16 | Q 31 | Page 122

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