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 ${sin}^{-1}x+{sin}^{-1}y+{sin}^{-1}z=3\pi /2$, then the value of

$x^{100}+y^{100}+z^{100}-\frac{9}{x^{101}+y^{101}+z^{101}}$ is

(b) ${sin}^{-1}\frac{ax}{c}+{sin}^{-1}\frac{bx}{c}={sin}^{-1}x$

where $a^2+b^2=c^2,c\ne 0$.

Answer 1

(a) . We know that $\left|{sin}^{-1}x\right|\le \pi /2$

Hence from the given relation we observe that each of ${sin}^{-1}x,{sin}^{-1}y$ and ${sin}^{-1}z$ will be $\pi /2$ so that $x=y=z=sin\left(\pi /2\right)=1$ $\therefore$ $3-\frac{9}{3}=0$

(b) ${sin}^{-1}[\frac{ax}{c}\sqrt{1-\frac{b^2{x}^2}{c^2}}+\frac{bx}{c}\sqrt{1-\frac{a^2{x}^2}{c^2}}]={sin}^{-1}x$

$\therefore$ $x[a\sqrt{c^2-b^{22}{x}^{}}+b\sqrt{c^2-a^2{x}^2}-c^2]=0$

$\therefore$ $x=0$ is one solution which satisfies the given equation.

Again $a\sqrt{c^2-b^2{x}^2}+b\sqrt{c^2-a^2{x}^2}=c^2$

Squaring both sides,

$a^2(c^2-b^2{x}^2)+b^2(c^2-a^2{x}^2)+2ab\sqrt{}\sqrt{}=c^4$

$c^2(a^2+{b2}^{})-2a^2{b}^2{x}^2+2ab\sqrt{}\sqrt{}=c^4$

Put $a^2+b^2=c^2$ and cancel $c^4$ from both sides

$\therefore$ $ab{x}^2=\sqrt{c^2-b^2{x}^2}\sqrt{c^2-a^2{x}^2}$

Square again

$a^2{b}^2{x}^4=c^4-c^2(a^2+b^2)x^2+a^2{b}^2{x}^4$

or $0=c^4(1-x^2)\therefore x=\pm 1$