Question

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

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

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

solve for x the following equations :

(a) ${cot}^{-1}x+{sin}^{-1}\frac{1}{\surd \left(5\right)}=\frac{\pi }{4}$

(b) $2{tan}^{-1}\left(cosx\right)=tan{tan}^{-1}\left(2cosecx\right)$.

(c) $tan\left({cos}^{-1}x\right)=sin\left({cot}^{-1}\frac{1}{2}\right)$

Answer 1

(a) ${sin}^{-1}\frac{1}{\surd \left(5\right)}={tan}^{-1}\frac{1}{2}$ and $tan\frac{\pi }{4}=1$

Ans. $x=3$

(b) The given equation is equivalent to the equation

${tan}^{-1}\frac{2cosx}{1-{cos}^{2}x}={tan}^{-1}\left(2cosecx\right)$

or $2cosx.{cosec}^{2}x=2cosecx$

or $cosecx(cotx-1)=0$

$\therefore$ $cotx-1=0$ or $cotx=1$ $[\because cosecx\ne 0]$

$\therefore$ $x=n\pi +\pi /4$, $nϵI$.

(c) We write the given equation as

$tan\left[{tan}^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)\right]=sin\left({sin}^{-1}\frac{2}{\surd \left(5\right)}\right)$

or $\frac{\sqrt{1-x^2}}{x}=\frac{2}{\surd \left(5\right)}$

or $5(1-x^2)=4x^2$

or $9x^2=5$, $\therefore x=\pm \left(\sqrt{5}/3\right)$