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) Solve the equation :
${cos}^{-1}\left(\sqrt{6}x\right)+{cos}^{-1}\left(3\sqrt{3}{x}^2\right)=\frac{\pi }{2}$
(b) If ${sin}^{1}6x+{sin}^{-1}6\sqrt{3}x=-\pi /2$, then $x=.......$
(c) ${sin}^{-1}x+{sin}^{-1}2x=\pi /3$

Answer 1

(a) ${cos}^{-1}3\sqrt{3}{x}^2=\frac{\pi }{2}-{cos}^{-1}\left(\sqrt{6}x\right)={sin}^{-1}\left(\sqrt{6}x\right)={cos}^{-1}(1-6x^2)$

$\therefore$ $3\sqrt{3}{x}^2=1+6x^2$

or $x^2(6+3\sqrt{3})=1$

$\therefore$ $x^2=\frac{1}{6+3\sqrt{3}}=\frac{6-3\sqrt{3}}{9}=\frac{2-\sqrt{3}}{3}=\frac{4-2\sqrt{3}}{6}$

or $x^2=\frac{{(2-\sqrt{3})}^2}{6}$ $\therefore$ $x=\frac{2-\sqrt{3}}{\sqrt{6}}$

(b) Ans. $-\frac{1}{12}$

$\frac{\pi }{2}+{sin}^{-1}6\sqrt{3}x=-{sin}^{-1}6x$. Take sine.

$cos\left({sin}^{-1}6\sqrt{3}x\right)=-6x$

or $\sqrt{1-108x^2}=-6x$. Square.

$1-108x^2=36x^2$ or $x^2=1/144$

$\therefore$ $x=\frac{1}{12},-\frac{1}{12}$

Since we have squared we must check the roots. Clearly $x=-\frac{1}{12}$ satisfies and $x=\frac{1}{12}$ does not satisfy.

(c) ${sin}^{-1}x+{sin}^{-1}2x=\pi /3={sin}^{-1}\left(\sqrt{3}/2\right)$

$\therefore$ ${sin}^{-1}x-{sin}^{-1}\left(\sqrt{3}/2\right)=-{sin}^{-1}2x$

$\therefore$ ${sin}^{-1}[x\sqrt{(1-\frac{3}{4})}-\frac{\sqrt{3}}{2}\sqrt{1-x^2}]=-{sin}^{-1}2x$

or $\frac{x}{2}-\frac{\sqrt{3}}{2}\sqrt{1-x^2}=-2x$.

or $5x=\sqrt{3}\sqrt{1-x^2}$. Square

$25x^2=3-3x^2$ or $28x^2=3$,

$\therefore$ $x=\pm \sqrt{3}/\left(2\sqrt{7}\right)$

Clearly from the given equation the value of x must be $+ive$

$\therefore$ $x=\frac{1}{2}\sqrt{\frac{3}{7}}$