Question

In triangle $A B C, \angle C=\frac{\pi}{2}$ and $\sin ^{-1}(x)=\sin ^{-1}\left(\frac{a x}{c}\right)+\sin ^{-1}\left(\frac{b x}{c}\right)$, 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, $|x| \leq 1$, also $\frac{a^{2} x^{2}}{c^{2}}+\frac{b^{2} x^{2}}{c^{2}}=x^{2} \leq 1$
$\sin ^{-1} x=\sin ^{-1}\left(\frac{a x}{c}\right)+\sin ^{-1}\left(\frac{b x}{c}\right)$
$\Rightarrow \sin ^{-1} x=\sin ^{-1}\left(\frac{a x}{c} \sqrt{1-\frac{b^{2} x^{2}}{c^{2}}}+\frac{b x}{c} \sqrt{1-\frac{a^{2} x^{2}}{c^{2}}}\right)$
$\Rightarrow x=\frac{a x}{c^{2}} \sqrt{c^{2}-b^{2} x^{2}}+\frac{b x}{c^{2}} \sqrt{c^{2}-a^{2} x^{2}}$
$\Rightarrow x=0$ or $c^{2}=a \sqrt{c^{2}-b^{2} x^{2}}+b \sqrt{c^{2}-a^{2} x^{2}}$
$\Rightarrow c^{4}=a^{2}\left(c^{2}-b^{2} x^{2}\right)+b^{2}\left(c^{2}-a^{2} x^{2}\right)$
$+2 a b \sqrt{\left(c^{2}-b^{2} x^{2}\right)\left(c^{2}-a^{2} x^{2}\right)}$
$\Rightarrow c^{4}=\left(a^{2}+b^{2}\right) c^{2}-2 a^{2} b^{2} x^{2}+2 a b \sqrt{\left(c^{2}-b^{2} x^{2}\right)\left(c^{2}-a^{2} x^{2}\right)}$
$\Rightarrow a b 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 $x=0,1,-1$