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If $y=\sqrt{\left(\frac{1+\cos 2 \theta}{1-\cos 2 \theta}\right)}$, then $\frac{d y}{d \theta}$ at $\theta=\frac{3 \pi}{4}$ is :
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The correct answer is:
-2
$y=\sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}$
$\Rightarrow y=\sqrt{\frac{2 \cos ^2 \theta}{2 \sin ^2 \theta}}=\sqrt{\cot ^2 \theta}$
$\Rightarrow y=\cot \theta$
Differentiate w.r.t. ' $\theta$ ', we get: $\frac{d y}{d \theta}=-\operatorname{cosec}^2 \theta$
Now, $\left(\frac{\mathrm{dy}}{\mathrm{d} \theta}\right)_{\theta=\frac{3 \pi}{4}}=-\operatorname{cosec}^2\left(\frac{3 \pi}{4}\right)$
$=-\operatorname{cosec}^2\left(\pi-\frac{\pi}{4}\right)=-\operatorname{cosec}^2 \frac{\pi}{4}=-2$
$\Rightarrow y=\sqrt{\frac{2 \cos ^2 \theta}{2 \sin ^2 \theta}}=\sqrt{\cot ^2 \theta}$
$\Rightarrow y=\cot \theta$
Differentiate w.r.t. ' $\theta$ ', we get: $\frac{d y}{d \theta}=-\operatorname{cosec}^2 \theta$
Now, $\left(\frac{\mathrm{dy}}{\mathrm{d} \theta}\right)_{\theta=\frac{3 \pi}{4}}=-\operatorname{cosec}^2\left(\frac{3 \pi}{4}\right)$
$=-\operatorname{cosec}^2\left(\pi-\frac{\pi}{4}\right)=-\operatorname{cosec}^2 \frac{\pi}{4}=-2$
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