Question

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

• A. 2
• B. 3
• C. 4
• D. None

Answer 1

The correct option is B 3

$\angle C=\frac{\pi }{2}\Rightarrow a^2+b^2=c^2$.
Clearly $|x|\le 1$ and $\frac{a^2{x}^2}{c^2}+\frac{a^2{x}^2}{c^2}=x^2\le 1$
From the given equation,
$\Rightarrow x=\frac{ax}{c}\sqrt{1-\frac{b^2{x}^2}{c^2}}+\frac{bx}{c}\sqrt{1-\frac{a^2{x}^2}{c^2}}$
$\Rightarrow x=0$ or $c^2=a\sqrt{c^2-b^2{x}^2}+b\sqrt{c^2-a^2{x}^2}$
$\Rightarrow x=0$ or $c^4=a^2(c^2-b^2{x}^2)+b^2(c^2-a^2{x}^2)$
$+2ab\sqrt{c^2-b^2{x}^2}\sqrt{c^2-a^2{x}^2}$
$\Rightarrow a^2{b}^2{x}^4=c^4-(a^2+b^2)c^2{x}^2+a^2{b}^2{x}^4$ or x=0
$\Rightarrow c^4=c^4{x}^2$ or x=0 $\Rightarrow x=0$ or $\pm 1$
Hence, total number of different values of x are three; x = 0, - 1, 1. It is easy to see that all these values satisfy the equation.