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 ${tan}^{-1}(x+\frac{2}{x})-{tan}^{-1}\frac{4}{x}-{tan}^{-1}(x-\frac{2}{x})=0$

then $x=..........$ or $x=........\left(xϵR\right)$

(b) If ${sin}^{-1}x+{sin}^{-1}$ $y=2\pi /3$

${cos}^{-1}x-{cos}^{-1}$ $y=\pi /3$

then $x=........$, $y=.......$

(c) It ${tan}^{-1}y=4{tan}^{-1}x,(\left|x\right|<tan\frac{\pi }{8})$, then

express y as an algebraic function of x, Also deduce that $tan\left(\pi /8\right)$ is a root of $x^4-6x^2+1=0$.

Answer 1

(a) $\pm \sqrt{2}$ combine 1st and 3rd.

$x^4=4$ $\therefore$ $x^2=2$ only as $-2$ is rejected.

(b) $x=1/2,y=1$. second relation can be written as

$(\frac{\pi }{2}-{sin}^{-1}x)-(\frac{\pi }{2}-{sin}^{-1}y)=\frac{\pi }{3}$

or ${sin}^{-1}y=\pi /2$, ${sin}^{-1}X=\pi /6$ etc.

(c) $2{tan}^{-1}x={tan}^{-1}\frac{2x}{1-x^2}$

$\therefore$ $4{tan}^{-1}x=2{tan}^{-1}\frac{2x}{1-x^2}$

$4{tan}^{-1}x={tan}^{-1}\frac{2.\frac{2x}{1-x^2}}{1-{\left(\frac{2x}{1-x^2}\right)}^2}$

or ${tan}^{-1}y={tan}^{-1}\frac{4x(1-x^2)}{x^4-6x^2+1}$

$\therefore$ $y=\frac{4x(1-x^2)}{x^4-6x^2+1}$

Now if $x=tan\frac{\pi }{8}$, then

${tan}^{-1}x=\frac{\pi }{8}$ or $4{tan}^{-1}x=\frac{\pi }{2}={tan}^{-1}y$

$\therefore$ $y=tan\frac{\pi }{2}=\infty$

$\therefore$ $D^r$ of $y=0$ or $x^4-6x^2+1=0$

where $x=tan\frac{\pi }{8}$ is a root of above.