Question
$\frac{d}{{dx}}\left\{ {{{\cos }^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)} \right\} = $
माना $\frac{{1 - {x^2}}}{{1 + {x^2}}} = \cos \theta $
==> $1 - {x^2} = (1 + {x^2})\cos \theta $
==> $ - {x^2}(1 + \cos \theta ) = \cos \theta - 1$
==> ${x^2} = \frac{{1 - \cos \theta }}{{1 + \cos \theta }} = \frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2{{\cos }^2}\frac{\theta }{2}}} = {\tan ^2}\frac{\theta }{2}$
या $x = \tan \frac{\theta }{2}$ या $\theta = 2{\tan ^{ - 1}}x$
अत: $\frac{d}{{dx}}(\theta ) = \frac{d}{{dx}}(2{\tan ^{ - 1}}x) = \frac{2}{{1 + {x^2}}}$.
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