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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)`
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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