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Mathematics Question on Continuity and differentiability

If y=(1+cos⁡2θ1−cos⁡2θ)y = \sqrt{\left(\frac{1+\cos 2\theta}{1 -\cos 2\theta}\right)} , then dydθ\frac{dy}{d\theta} at θ=3π4\theta =\frac{3\pi}{4} is:

A

-2

B

2

C

±\pm

D

none of these

Answer

-2

Explanation

Solution

y=(1+cos⁡2θ1−cos⁡2θ)y = \sqrt{\left(\frac{1+\cos 2\theta}{1 -\cos 2\theta}\right)} , ⇒  y=2 cos⁡2θ2 sin⁡2θ=cot⁡2 θ\Rightarrow \:\: y = \sqrt{ \frac{2 \:\cos^2 \theta}{2 \: \sin^2\theta}} = \sqrt{\cot^2 \: \theta} ⇒  y=cot⁡ θ\Rightarrow \:\: y = \cot \: \theta Differentiate w.r.t. ′θ′' \theta ', we get dydθ=−cosec2θ\frac{dy }{d \theta} = - cosec^2 \theta Now , (dydθ)θ=3π4 \left( \frac{dy }{d \theta}\right)_{\theta = \frac{3 \pi}{4}} =−cosec2(3π4) = -cosec^2 \left( \frac{3 \pi}{4} \right) =−cosec2(π−π4) = -cosec^2 \left(\pi - \frac{ \pi}{4} \right) =−cosec2π4 = -cosec^2 \frac{ \pi}{4} =−2(∵sin⁡π4=12)= - 2 \left( \because \sin \frac{ \pi}{4} = \frac{ 1}{\sqrt{2}} \right)