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If $y=\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$, then the values of $\frac{d y}{d x}$ and $\frac{d^2 y}{d x^2}$ respectively are
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Verified Answer
The correct answer is:
$\frac{1}{2}, 0$
We have
$y=\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$
$\Rightarrow \quad y=\tan ^{-1} \sqrt{\frac{2 \sin ^2 x / 2}{2 \cos ^2 x / 2}}=\tan ^{-1}(\tan x / 2)$
$\Rightarrow \quad y=x / 2 \Rightarrow \frac{d y}{d x}=\frac{1}{2}$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=0$
$y=\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}$
$\Rightarrow \quad y=\tan ^{-1} \sqrt{\frac{2 \sin ^2 x / 2}{2 \cos ^2 x / 2}}=\tan ^{-1}(\tan x / 2)$
$\Rightarrow \quad y=x / 2 \Rightarrow \frac{d y}{d x}=\frac{1}{2}$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=0$
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