The value of the limit \mathop \lim \limits_\theta \to 0 \tan (\pi \cos ^2\theta ) \over \sin (2\pi \sin ^2\theta ) is equal to :

Limits Continuity and Differentiability PYQ — JEE Main 2021 · 17 Mar · Shift 2 · Q29 with Solution

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Correct answer: B

We use $\cos^2\theta = 1-\sin^2\theta.$ So, $\tan(\pi \cos^2\theta)=\tan\big(\pi(1-\sin^2\theta)\big).$

Using the identity $\tan(\pi-x)=-\tan x,$ we get $\tan(\pi \cos^2\theta)= -\tan(\pi \sin^2\theta).$

Hence the limit becomes $L=\lim_{\theta\to 0}\frac{-\tan(\pi\sin^2\theta)}{\sin(2\pi\sin^2\theta)}.$

Let $x=\pi\sin^2\theta.$ As $\theta\to 0$ , we have $\sin^2\theta\to 0$ , so $x\to 0$ .

Then $L=\lim_{x\to 0}\frac{-\tan x}{\sin 2x}.$

Now use

\qquad \tan x = \frac{\sin x}{\cos x}.$$ So, $$\frac{-\tan x}{\sin 2x} = -\frac{\sin x/\cos x}{2\sin x\cos x} = -\frac{1}{2\cos^2 x}.$$ 3. **Take the limit** As $x\to 0$, $$\cos x \to 1 \quad \Rightarrow \quad \cos^2 x \to 1.$$ Therefore, $$L= -\frac{1}{2\cdot 1}= -\frac{1}{2}.$$ 4. **Check the options** The correct option is: $$\boxed{\text{B }\left(-\frac12\right)}.$$