Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

${\sin }^{-1}\frac{3x}{5}+{\sin }^{-1}\frac{4x}{5}={\sin }^{-1}x$

${\sin }^{-1}\left(\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}\right)={\sin }^{-1}x$

$\frac{3x}{5}\sqrt{1-\frac{16x^2}{25}}+\frac{4x}{5}\sqrt{1-\frac{9x^2}{25}}=x$

$x=0, 3\sqrt{25-16x^2}+4\sqrt{25-9x^2}=25$

$4\sqrt{25-9x^2}=25-3\sqrt{25-16x^2}$ 

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 $=3$