Question

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

Answer 1

Let $x=cos\theta$ so that $co{s}^{-1}x=\theta$
$\therefore ta{n}^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=ta{n}^{-1}\left(\frac{\sqrt{1+cos\theta }-\sqrt{1-cos\theta }}{\sqrt{1+cos\theta }+\sqrt{1-cos\theta }}\right)=ta{n}^{-1}\left(\frac{\sqrt{2}cos\frac{\theta }{2}-\sqrt{2}sin\frac{\theta }{2}}{\sqrt{2}cos\frac{\theta }{2}+\sqrt{2}sin\frac{\theta }{2}}\right)(\because 1+cos\theta =2co{s}^2\frac{\theta }{2}and1-cos\theta =2si{n}^2\frac{\theta }{2})=ta{n}^{-1}\left(\frac{cos\frac{\theta }{2}-sin\frac{\theta }{2}}{cos\frac{\theta }{2}+sin\frac{\theta }{2}}\right)=ta{n}^{-1}\left(\frac{1-tan\frac{\theta }{2}}{1+tan\frac{\theta }{2}}\right)$
(inside the bracket divide numerator and denominator by $cos\frac{\theta }{2})$
$=ta{n}^{-1}\left(tan(\frac{\pi }{4}-\frac{\theta }{2})\right)[\because tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}]=\frac{\pi }{4}-\frac{\theta }{2}=\frac{\pi }{4}-\frac{1}{2}co{s}^{-1}x$