Question

Solve :
${\cos }^{-1}\left(\frac{a^{-x}-a^x}{a^{-x}+a^x}\right)$

Answer 1

$\begin{array}{c}{\cos }^{-1}\left(\frac{\frac{1}{a^x}-a^x}{\frac{1}{a^x}+a^x}\right)\\ ={\cos }^{-1}\left(\frac{1-{\left(a^x\right)}^2}{1+{\left(a^x\right)}^2}\right)\\ Now,\\ put{a}^x=\tan \theta \\ \Rightarrow \theta ={\tan }^{-1}{a}^x\\ \Rightarrow 2\theta =2{\tan }^{-1}{a}^x\\ Then,\\ {cos}^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)\\ ={\cos }^{-1}\left(\cos 2\theta \right)\left(\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)\\ =2\theta \\ But\\ 0\le 2\theta <\pi \\ 0\le \theta <\frac{\pi }{2}\\ 0\le {\tan }^{-1}{a}^x<\frac{\pi }{2}\\ 0\le a^x<\infty \\ Hence,solved.\end{array}$