Write the function in the simplest form: ^ - 1 1 + x^2 - 1 x ,x 0… - Vidyadip

Question

Write the function in the simplest form: ${\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x},x \ne 0$.

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Answer

Put $x = \tan \theta$ hence $\theta = {\tan ^{ - 1}}x$
Now, $ {\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}} - 1}}{x}$
$ = {\tan ^{ - 1}}\frac{{\sqrt {1 + {{\tan }^2}\theta } - 1}}{{\tan \theta }}$
$ = {\tan ^{ - 1}}\frac{{\sec \theta - 1}}{{\tan \theta }}$
$ = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{{\cos \theta }} - 1}}{{\frac{{\sin \theta }}{{\cos \theta }}}}} \right)$
$= {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)$
$ = {\tan ^{ - 1}}\left( {\tan \frac{\theta }{2}} \right)$
$= \frac{\theta }{2} = \frac{1}{2}{\tan ^{ - 1}}x$

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