Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
Question
$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ then x is equal to
✓
Answer
$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ Now, we will put x = sin y in the given equation, and we get $\sin ^{-1}(1-\sin y)-2 \sin ^{-1} \sin y=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1}(1-\sin y)-2 y=\frac{\pi}{2}$ $\Rightarrow \sin ^{-1}(1-\sin y)=\frac{\pi}{2}+2 y$ $\Rightarrow 1-\sin y=\sin \left(\frac{\pi}{2}+2 y\right)$ $\Rightarrow 1-\sin y=\cos 2 y \text ({ as } \sin \left(\frac{\pi}{2}+x\right)=\cos x)$ $\Rightarrow$ $1 - cos 2y = sin y$ $\Rightarrow$ $2~sin2 y = sin y$ $\Rightarrow$ $sin y .(2 sin y - 1) = 0$ $\Rightarrow$ $sin y = 0 ~~or~~siny= \frac 12 $ $\therefore$ $x = 0~~ or~~x=\frac 12 $ Now, if we put x = $\frac{1}{2}$, then we will see that, L.H.S. = $\sin ^{-1}\left(1-\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$ = $\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$ = $-\sin ^{-1} \frac{1}{2}$ = $-\frac{\pi}{6} \neq \frac{\pi}{2} \neq$ R.H.S Hence, x = $\frac{1}{2}$ is not the solution of the given equation. Thus, x = 0
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