Question

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy ${\sin }^{-1}\left(\frac{3x}{5}\right)+{\sin }^{-1}\left(\frac{4x}{5}\right)={\sin }^{-1}x$ is equal to :

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is C $3$

${\sin }^{-1}\left(\frac{3x}{5}\right)+{\sin }^{-1}\left(\frac{4x}{5}\right)={\sin }^{-1}x$
Taking sine on both sides
$\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}=x$
$\Rightarrow 3x\sqrt{25-16x^2}=25x-4x\sqrt{25-9x^2}$
$\Rightarrow x=0$
or $3\sqrt{25-16x^2}=25-4\sqrt{25-9x^2}$
$\Rightarrow 9(25-16x^2)=625-200\sqrt{25-9x^2}+16(25-9x^2)$
$\Rightarrow 200\sqrt{25-9x^2}=800$
$\Rightarrow \sqrt{25-9x^2}=4$
$\Rightarrow x^2=1$
$\Rightarrow x=\pm 1$
$\therefore$ Total number of solutions $=3$