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 :

Given,<br><br>$$\mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi {{\cos }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$<br><br>$$ = \mathop {\lim }\limits_{\theta \to 0} {{\tan (\pi - \pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $\therefore$ $\left( {{{\cos }^2}\theta = 1 - {{\sin }^2}\theta } \right)$<br><br>$$ = \mathop {\lim }\limits_{\theta \to 0} {{ - \tan (\pi {{\sin }^2}\theta )} \over {\sin (2\pi {{\sin }^2}\theta )}}$$ $\therefore$ $(\tan (\pi - \theta ) = - \tan \theta )$<br><br>$$ = \mathop {\lim }\limits_{\theta \to 0} {{{{ - \tan (\pi {{\sin }^2}\theta )} \over {\pi {{\sin }^2}\theta }}} \over {{{\sin (2\pi {{\sin }^2}\theta )} \over {2\pi {{\sin }^2}\theta }} \times 2}}\left( \matrix{ As\,\theta \to 0 \hfill \cr then \,{\sin ^2}\theta \to 0 \hfill \cr} \right)$$<br><br>$= -{1 \over 2}.$ $\because$ $$\left( \matrix{ \mathop {\lim }\limits_{\theta \to 0} {{\tan \theta } \over \theta } \to 1 \hfill \cr \&amp; \,\mathop {\lim }\limits_{\theta \to 0} {{\sin \theta } \over \theta } = 1 \hfill \cr} \right)$$