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  • Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 15

Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 15

Answers (1)

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Answer:

    -1

Hint:

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

Given:

y=\tan ^{-1}\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right)

Solution:  

\begin{aligned} &y=\tan ^{-1}\left(\sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}}\right) \\\\ &y=\tan ^{-1}\left(\sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}}\right) \end{aligned}

Using

\begin{aligned} &1-\cos 2 x=2 \sin ^{2} x \\\\ &1+\cos 2 x=2 \cos ^{2} x \\\\ &y=\tan ^{-1}\left(\sqrt{\tan ^{2} x}\right) \end{aligned}

\begin{aligned} &y=\tan ^{-1}(-\tan x)\left[-\frac{\pi}{2}<x<0\right] \\\\ &y=-x \quad\left[\because \tan ^{-1}(\tan x)=x\right] \end{aligned}

Differentiate it w.r.t x, we get

\begin{aligned} &\frac{d y}{d x}=\frac{d(-x)}{d x} \\\\ &\frac{d y}{d x}=-1 \text { for } x \in\left(-\frac{\pi}{2}, 0\right) \end{aligned}

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