Question

Prove that $ta{n}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}{x}^2.$

Answer 1

We have,

$ta{n}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}{x}^2$

$\therefore LHS=ta{n}^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)\cdots \left(i\right)[let{x}^2=cos2\theta =(co{s}^2\theta -si{n}^2\theta )=1-2si{n}^2\theta =2co{s}^2\theta -1]$

$\Rightarrow co{s}^{-1}{x}^2=2\theta \Rightarrow \theta =\frac{1}{2}co{s}^{-1}{x}^2\therefore \sqrt{1+x^2}=\sqrt{1+cos2\theta }=\sqrt{1+2co{s}^2\theta -1}=\sqrt{2cos\theta }and\sqrt{1-x^2}=\sqrt{1-cos2\theta }=\sqrt{1-1+2si{n}^2\theta }=\sqrt{2}sin\theta \therefore LHS=ta{n}^{-1}\left(\frac{\sqrt{2}cos\theta +\sqrt{2}sin\theta }{\sqrt{2}cos\theta -\sqrt{2}sin\theta }\right)=ta{n}^{-1}\left(\frac{cos\theta +sin\theta }{cos\theta -sin\theta }\right)=ta{n}^{-1}\left(\frac{1+tan\theta }{1-tan\theta }\right)=ta{n}^{-1}\left(\frac{tan\frac{\pi }{4}+tan\theta }{1-tan\frac{\pi }{4}tan\theta }\right)=ta{n}^{-1}\left[tan(\frac{\pi }{4}+\theta )\right][\because tan(x+y)=\frac{tanx+tany}{1-tanx-tany}]=\frac{\pi }{4}+\theta =\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}{x}^2=RHSHenceproved.$