Write the following functions in the simplest form:

    5. \tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x},\;\;x\neq 0

Answers (1)

We have \tan^{-1}\frac{\sqrt{1 + x^2}- 1}{x}

Take x= \tan \Theta \Rightarrow \Theta = tan^{-1}x

\therefore \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{1-cos \Theta}{sin \Theta} \right )

=\tan^{-1}\left ( \frac {2sin^2\left ( \frac{\Theta}{2} \right )}{2sin\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 

=\frac{1}{2}\tan^{-1}x  is the simplified form.

 

 

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