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) $3{sin}^{-1}\frac{2x}{1+x^2}-4{cos}^{-1}\frac{1-x^2}{1+x^2}+2{tan}^{-1}\frac{2x}{1-x^2}=\frac{\pi }{3}$
(b) $sin\left[\right(1/5\left){cos}^{-1}x\right]=1$
(c) ${tan}^{-1}\sqrt{x^2+x}+{sin}^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$

Answer 1

(a) We have
$3\left(2{tan}^{-1}x\right)-4\left(2{tan}^{-1}x\right)+2\left(2{tan}^{-1}x\right)=\frac{\pi }{3}$.
or $2{tan}^{-1}x=\frac{\pi }{3}$ $\therefore$ ${tan}^{-1}x=\frac{\pi }{6}$
or $x=tan\left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}$.
(b) Let $\frac{1}{5}{cos}^{-1}x=\alpha$
Since $0\le {cos}^{-1}x\le \pi$,
we have $0\le \alpha \le \frac{\pi }{5}$
Hence $sin\alpha \ne 1$, that is, $sin\left(\frac{1}{5}{cos}^{-1}x\right)=1$ has no solution,
(c) Above equation is defined if $x^2+x\ge 0$
and $x^2+x+1\ge 0$ but $\le 1$
i.e., $0\le x^2+x+1\le 1$
Above is possible only when $x^2+x=0$
or $x(x+1)=0$ $\therefore$ $x=0,-1$.