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  • Prove that tan^{-1}( √1+x^{2}+√1-x^{2}/√1+x^{2}-√1-x^{2})=π/4+1/2 cos^{-1}x^{2}

Prove that \tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}

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We have \tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )=\frac{\pi}{4}+\frac{1}{2}cos^{-1}x^{2}

LHS=\tan^{-1}\left ( \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \right )........(i)

\left [ x^{2}=\cos2\theta=\cos^{2}\theta+\sin^{2}\theta=1-2\sin^{2}\theta = 2\cos^{2}\theta-1 \right ]

\Rightarrow \cos^{-1}x^{2}=2\theta

\Rightarrow \theta=\frac{1}{2}\cos^{-1}x^{2}

\therefore \sqrt{1+x^{2}}=\sqrt{1+\cos2\theta}

\Rightarrow \sqrt{1+2\cos^{2}\theta-1}=\sqrt{2}\cos \theta

And  \sqrt{1-x^{2}}=\sqrt{1-\cos2\theta}

\sqrt{1-1+2 \sin^{2}\theta}=\sqrt{2}\sin \theta

\therefore LHS = \tan^{-1}\left (\frac{ \sqrt{2}\cos \theta +\sqrt{2} \sin \theta}{ \sqrt{2}\cos \theta -\sqrt{2} \sin \theta}\right )

= \tan^{-1}\left (\frac{ \cos \theta + \sin \theta}{ \cos \theta - \sin \theta}\right )

= \tan^{-1}\left (\frac{1+\tan \theta}{ 1-\tan \theta}\right )

= \tan^{-1}\left \{\frac{\tan\left (\frac{\pi}{4} \right )+\tan \theta}{ \tan\left (\frac{\pi}{4} \right )-\tan \theta} \right \}

= \tan^{-1}\left [ \tan\left ( \frac{\pi}{4}+\theta \right ) \right ]  \left [ Since , \tan\left ( x+y \right )=\frac{\tan x+\tan y}{1-\tan x.\tan y} \right ]

=\frac{\pi}{4}+\theta

=\frac{\pi}{4}+\frac{1}{2}\cos^{-1}x^{2}

=RHS

LHS=RHS

Hence Proved

 

 

 

Posted by

infoexpert24

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