Question

Inverse circular functions,Principal values of ${\sin }^{-1}x,{cos}^{-1}x,{\tan }^{-1}x$.

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

$\pi +{\tan }^{-1}\frac{x+y}{1-xy}$, $xy>1$.

(a) If ${\sin }^{-1}\frac{2p}{1+p^2}-{\cos }^{-1}\frac{1-q^2}{1+q^2}={\tan }^{-1}\frac{2x}{1-x^2}$

then prove that $x=\frac{p-q}{1+pq}$.

(b) Solve for x

${\sin }^{-1}\frac{2a}{1+a^2}+{\sin }^{-1}\frac{2b}{1+b^2}=2{\tan }^{-1}x$

(c) Prove that

$\tan [\frac{1}{2}{\sin }^{-1}\frac{2a}{1+a^2}+\frac{1}{2}{\cos }^{-1}\frac{1-a^2}{1+a^2}]=\frac{2a}{1-a^2}$.

Answer 1

(a) The given equation is

$2{\tan }^{-1}p-2{\tan }^{-1}q=2{\tan }^{-1}x$

or ${\tan }^{-1}p-{\tan }^{-1}q={\tan }^{-1}x$

or ${\tan }^{-1}\frac{p-q}{1+pq}={tan}^{-1}x$

$\therefore$ $x=\frac{p-q}{1+pq}$.

(b) $x=\frac{a+b}{1-ab}$ As in part (a).

(c) L.H.S.$=\tan [\frac{1}{2}.2{\tan }^{-1}a+\frac{1}{2}.2{\tan }^{-1}a]$

$=\tan \left[2{\tan }^{-1}a\right]$

$=\tan {\tan }^{-1}\frac{2a}{1-a^2}=\frac{2a}{1-a^2}$