The value of the limit $\lim _{\theta \rightarrow 0} \frac{\tan \left(\pi \cos ^{2} \theta\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$ is equal to :
Correct Option: 1
$\lim _{\theta \rightarrow 0} \frac{\tan \left(\pi\left(1-\sin ^{2} \theta\right)\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$
$=\lim _{\theta \rightarrow 0} \frac{-\tan \left(\pi \sin ^{2} \theta\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$
$=\lim _{\theta \rightarrow 0}-\left(\frac{\tan \left(\pi \sin ^{2} \theta\right)}{\pi \sin ^{2} \theta}\right)\left(\frac{2 \pi \sin ^{2} \theta}{\sin \left(2 \pi \sin ^{2} \theta\right)}\right) \times \frac{1}{2}$
$=\frac{-1}{2}$
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