#### Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise Multiple Choice Questions Question 10 Maths Textbook Solution.

Hint: The function range should fulfill the interval.

Given: $\sqrt{1+\cos 2 x}=\sqrt{2} \sin ^{-1}(\sin x)$

Solution:

$-\pi \leq x \leq \frac{-\pi}{2}$

For,

\begin{aligned} &\sqrt{1+\cos 2 x}=\sqrt{2} \sin ^{-1}(\sin x) \\ &\sqrt{2}|\cos x|=\sqrt{2} \sin ^{-1}(\sin x) \\ &\sqrt{2}(-\cos x)=\sqrt{2}(-\pi-x) \\ &\cos x=\pi+x \end{aligned}

It doesn’t satisfy for any value of x in interval $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

For, $\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$

\begin{aligned} &\sqrt{1+\cos 2 x}=\sqrt{2} \sin ^{-1}(\sin x) \\ &\sqrt{2}|\cos x|=\sqrt{2} x \\ &\sqrt{2}(\cos x)=\sqrt{2} x \\ &\cos x=x \end{aligned}

It gives one value of x in interval $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

For, $\frac{\pi}{2} \leq x \leq \pi$

\begin{aligned} &\sqrt{1+\cos 2 x}=\sqrt{2} \sin ^{-1}(\sin x) \\ &\sqrt{2}|\cos x|=\sqrt{2} \sin ^{-1}(\sin x) \\ &\sqrt{2}(-\cos x)=\sqrt{2}(\pi-x) \\ &\cos x=-\pi+x \end{aligned}

It gives two real solutions on interval  $\left(\frac{\pi}{2}, \pi\right)$

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