Question

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

OR If $ta{n}^{-1}\left(\frac{x-2}{x-4}\right)+ta{n}^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi }{4}$, find the value of x.

Answer 1

LHS: Let $Y=ta{n}^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

Put $x=cos2\theta \Rightarrow \theta =\frac{1}{2}co{s}^{-1}x\dots \left(i\right)$

Let $Y=ta{n}^{-1}\left(\frac{\sqrt{1+cos2\theta }-\sqrt{1-cos2\theta }}{\sqrt{1+cos2\theta }+\sqrt{1-cos2\theta }}\right)=ta{n}^{-1}\left(\frac{\sqrt{2}cos\theta -\sqrt{2}sin\theta }{\sqrt{2}cos\theta +\sqrt{2}sin\theta }\right)$

$=ta{n}^{-1}\left(\frac{cos\theta -sin\theta }{cos\theta +sin\theta }\right)=ta{n}^{-1}\left(\frac{1-tan\theta }{1+tan\theta }\right)=ta{n}^{-1}\left(\frac{tan\frac{\pi }{4}-tan\theta }{1+tan\frac{\pi }{4}-tan\theta }\right)$

$=tan-1tan(\frac{\pi }{4}-\theta )=\frac{\pi }{4}-\theta$

$=\frac{\pi }{2}-\frac{1}{2}co{s}^{-1}x$ [By (i)]

=RHS

OR

The given equation is $ta{n}^{-1}\left(\frac{x-2}{x-4}\right)+ta{n}^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi }{4}$

$\Rightarrow ta{n}^{-1}\left(\frac{x-2}{x-4}\right)=ta{n}^{-1}\left(\frac{x+2}{x+4}\right)$

$\Rightarrow ta{n}^{-1}\left(\frac{x-2}{x-4}\right)=ta{n}^{-1}\left(\frac{1-\frac{x+2}{x+4}}{1+1\times \frac{x+2}{x+4}}\right)=ta{n}^{-1}\left(\frac{2}{2(x+3)}\right)$

$\Rightarrow \frac{x-2}{x-4}=\frac{1}{x+3}\Rightarrow x^3+x-6=x-4$

$\Rightarrow x^2=2\therefore x=\pm \sqrt{2}$.