Question

In triangle $ABC,\angle C=\frac{\pi }{2}$ and ${\sin }^{-1}(x)={\sin }^{-1}\left(\frac{ax}{c}\right)+{\sin }^{-1}\left(\frac{bx}{c}\right)$, where $a,b,c$ are the sides of triangle, then total number of different values of $x$ is

• A. 2
• B. 3
• C. 1
• D. None of these

Answer 1

Answer: (A)

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