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# Prove that $tan^{-1}\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\frac{\pi}{4}+\frac{1}{2}cos^{-1}x^2$

Toolbox:
• $put \: x^2 = cos\theta$
• $\Rightarrow \theta = cos^{-1}x^2$
• $1+cos\theta = 2cos^2\large\frac{\theta}{2} \: \: ; \: \; 1-cos\theta=2sin^2\large\frac{\theta}{2}$
• $\large\frac{tanA+tanB}{1-tanAtanB}=tan(A+B)$
• $tan\large\frac{\pi}{4}=1$
L.H.S.
put $x^2=cos\theta$
$\Rightarrow \theta = cos^{-1}x^2$
$\sqrt{1+x^2}=\sqrt{1+cos\theta}=\sqrt{2cos^2\large\frac{\theta}{2}}=\sqrt{2}cos\large\frac{\theta}{2}$
$\sqrt{1-x^2}=\sqrt{1-cos\theta}=\sqrt{2sin^2\large\frac{\theta}{2}}=\sqrt{2}sin\large\frac{\theta}{2}$
$\large\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=\large\frac{\sqrt{2}cos\large\frac{\theta}{2}+\sqrt{2}sin\large\frac{\theta}{2}}{\sqrt{2}cos\large\frac{\theta}{2}-\sqrt{2}sin\large\frac{\theta}{2}}$
$tan^{-1}\large\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}=tan^{-1} \bigg[ \large\frac{\sqrt{1+cos\theta}+\sqrt{1-cos\theta}}{\sqrt{1+cos\theta}-\sqrt{1-cos\theta}}\bigg]$
$= tan^{-1} \bigg[ \large\frac{cos\frac{\theta}{2}+sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}-sin\large\frac{\theta}{2}} \bigg]$

Divide num.and den. by $cos\large\frac{\theta}{2}$
$=tan^{-1}\bigg[\large\frac{1+\large\frac{sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}{1-\large\frac{sin\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}\bigg]$=$tan^{-1}\big(\frac{1+tan\large\frac{\theta}{2}}{1-tan\large\frac{\theta}{2}}\big)$

By writing $1=tan\large\frac{\pi}{4}$ in the numerator we get
$tan^{-1}\bigg(\frac{tan\large\frac{\pi}{4}+tan\large\frac{\theta}{2}}{1-tan\large\frac{\pi}{4}.tan\large\frac{\theta}{4}}\bigg)$
By taking A =$\large\frac{\pi}{4}\:and\:B=\large\frac{\theta}{2}$ in the formula of tan(A+B). we get
$\large\frac{tan\large\frac{\pi}{4}+tan\large\frac{\theta}{2}}{1-tan\large\frac{\pi}{4}.tan\large\frac{\theta}{4}}=tan(\large\frac{\pi}{4}+\large\frac{\theta}{2})$
$\Rightarrow\: tan^{-1}tan \bigg( \large\frac{\pi}{4}+\large\frac{\theta}{2} \bigg) = \large\frac{\pi}{4}+\large\frac{\theta}{2}$

By substituting the value of $\theta=cos^{-1}x$ $\Rightarrow\:\large\frac{\pi}{4}+\large\frac{1}{2} cos^{-1}x$
=R.H.S

edited Mar 19, 2013