2. inverse trigonometric function — ગણિત ધોરણ 12 સાયન્સ — Question

$\Rightarrow-2 \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1}(1-x)$

$\Rightarrow-2 \sin ^{-1} x=\cos ^{-1}(1-x)$ $\dots(1)$

Let $\sin ^{-1} x=\theta \Rightarrow \sin \theta=x \Rightarrow \cos \theta=\sqrt{1-x^{2}}$

$\therefore \theta=\cos ^{-1}(\sqrt{1-x^{2}})$

$\therefore \sin ^{-1} x=\cos ^{-1}(\sqrt{1-x^{2}})$

Therefore, from equation $( 1 )$, we have

$-2 \cos ^{-1}(\sqrt{1-x^{2}})=\cos ^{-1}(1-x)$

Let, $x=\sin y .$ Then, we have:

$-2 \cos ^{-1}(\sqrt{1-\sin ^{2} y})=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 \cos ^{-1}(\cos y)=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 y=\cos ^{-1}(1-\sin y)$

$\Rightarrow 1-\sin y=\cos (-2 y)=\cos 2 y$

$\Rightarrow 1-\sin y=1-2 \sin ^{2} y$

$\Rightarrow 2 \sin ^{2} y-\sin y=0$

$\Rightarrow \sin y(2 \sin y-1)=0$

$\Rightarrow \sin y=0$ or $\frac{1}{2}$

$\therefore x=0$ OR $x=\frac{1}{2}$

When $x=\frac{1}{2},$ it can be observed that

L.H.S. $=\sin ^{-1}\left(1-\frac{1}{2}\right)-\sin ^{-1} \frac{1}{2}$

$=\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$

$=-\sin ^{-1} \frac{1}{2}$

$=\frac{\pi}{6} \neq \frac{\pi}{2} \neq R . H . S$

$\therefore x=\frac{1}{2}$ is not a solution of the given equation. Thus, $x=0$