Let $$S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$$.
If $\mathrm{n(S)}$ denotes the number of elements in $\mathrm{S}$ then :
Solution
$$ {\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)}$$
<br/><br/>$\begin{aligned} & \text { Put } x=\tan \theta \quad \theta \in\left(0, \frac{\pi}{4}\right) \\\\ & 2 \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\\\ & 2 \tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]=\cos ^{-1}[\cos (2 \theta)] \\\\ & \Rightarrow 2\left(\frac{\pi}{4}-\theta\right)=2 \theta \Rightarrow \theta=\frac{\pi}{8} \\\\ & \Rightarrow x=\tan \frac{\pi}{8}=\sqrt{2}-1 \simeq 0.414\end{aligned}$
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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