Question

In triangle $ABC,$ right angled at $C,$ the number of values of $x$ such that ${\sin }^{-1}\left(x\right)={\sin }^{-1}\left(\frac{ax}{c}\right)+{\sin }^{-1}\left(\frac{bx}{c}\right)$, where $a,b,c$ are the sides of triangle is

Answer 1

$\angle C=\frac{\pi }{2}\Rightarrow a^2+b^2=c^2$
Clearly, $|x|\le 1.$
Also, $\frac{a^2{x}^2}{c^2}+\frac{b^2{x}^2}{c^2}=x^2\le 1$
${\sin }^{-1}x={\sin }^{-1}\left(\frac{ax}{c}\right)+{\sin }^{-1}\left(\frac{bx}{c}\right)$
$\Rightarrow {\sin }^{-1}x={\sin }^{-1}(\frac{ax}{c}\sqrt{1-\frac{b^2{x}^2}{c^2}}+\frac{bx}{c}\sqrt{1-\frac{a^2{x}^2}{c^2}})$
$\Rightarrow x=\frac{ax}{c^2}\sqrt{c^2-b^2{x}^2}+\frac{bx}{c^2}\sqrt{c^2-a^2{x}^2}$
$\Rightarrow x=0$
or $c^2=a\sqrt{c^2-b^2{x}^2}+b\sqrt{c^2-a^2{x}^2}$
$\Rightarrow c^4=a^2(c^2-b^2{x}^2)+b^2(c^2-a^2{x}^2)+2ab\sqrt{(c^2-b^2{x}^2)}\sqrt{(c^2-a^2{x}^2)}$
$\Rightarrow c^4=(a^2+b^2)c^2-2a^2{b}^2{x}^2+2ab\sqrt{(c^2-b^2{x}^2)(c^2-a^2{x}^2)}$
$\Rightarrow ab{x}^2=\sqrt{c^4-c^2(a^2+b^2)x^2+a^2{b}^2{x}^4}$
$\Rightarrow a^2{b}^2{x}^4=c^4-c^4{x}^2+a^2{b}^2{x}^4\Rightarrow x^2=1$
$\therefore x=1,-1$
Total number of different values of $x$ is three, namely $x=0,1,-1$