Question

Inverse circular functions,Principal values of ${\sin }^{-1}x,{\cos }^{-1}x,{\tan }^{-1}x$.
${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{\tan }^{-1}\frac{x+y}{1-xy}$, $xy>1$.
Prove that
${\tan }^{-1}\frac{1-x}{1+x}{\tan }^{-1}\frac{1-y}{1+y}={\sin }^{-1}\frac{y-x}{\sqrt{1+x^2}\sqrt{1+y^2}}$

Answer 1

Put $x=\tan \theta ,y=\tan \varphi$

$\therefore$ L.H.S.$=(\frac{\pi }{4}-\theta )-(\frac{\pi }{4}-\varphi )=\varphi -\theta$

$={\tan }^{-1}y{-\tan }^{-1}x={\tan }^{-1}\frac{y-x}{1+xy}$

$={\sin }^{-1}\frac{y-x}{{\{{(y-x)}^2+{(1+xy)}^2\}}^{1/2}}$

$={\sin }^{-1}\frac{y-x}{\surd \left\{\right(1+x^2\left)\right(1+y^2\left)\right\}}$

LHS=RHS

Hence proved.