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. 2
• B. 1
• C. 3
• D. 0

Answer 1

Answer: (C)

${\sin }^{-1}\frac{3x}{5}+{\sin }^{-1}\frac{4x}{5}={\sin }^{-1}x$
${\sin }^{-1}\left(\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}\right)={\sin }^{-1}x$
$\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}=x$
$x=0,3\sqrt{25-16x^2}+4\sqrt{25-9x^2}=25$
$4\sqrt{25-9x^2}=25-3\sqrt{25-16x^2}$ squaring we get
$16\left(25-9x^2\right)=625+9\left(25-16x^2\right)-150\sqrt{25-16x^2}$
$400=625+225-150\sqrt{25-16x^2}$
$\sqrt{25-16x^2}=3\Rightarrow 25-16x^2=9$
$\Rightarrow x^2=1$
Put $x=0,1,-1$ in the original equation We see that all values satisfy the original equation.
Number of solution $=3$