Explanation

(c)
Put the value of
$(x,\ y) \equiv (\tan{\theta + \sin{\theta,\tan{\theta - \sin{\theta)}}}}$ In the given option, we get the required result.
On putting the value of $x$ and $y$ in option (c), we get
$\lbrack({\tan{\theta + \sin{\theta)}}}^{2} - ({\tan{\theta - \sin{\theta^{2}\rbrack}}}^{2} = 16(\tan{\theta + \sin{\theta) \times (\tan{\theta - \sin{\theta)}}}}$
$\Rightarrow \lbrack\tan^{2}{\theta + \sin^{2}{\theta - \tan^{2}{\theta - \sin^{2}{\theta + 4\tan{\theta{\sin{\theta\rbrack}}^{2}}} = 16(\tan^{2}{\theta - \sin^{2}{\theta)}}}}}$
$\Rightarrow (4{\tan{\theta \bullet \sin{\theta)}}}^{2} = 16(\tan^{2}{\theta - \sin^{2}{\theta)}}$
$\Rightarrow 16\tan^{2}{\theta\sin^{2}{\theta = 16\tan^{2}{\theta\ (1 - \cos^{2}{\theta)}}}}$
$\Rightarrow \ \ 16\tan^{2}{\bullet \theta}\sin^{2}{\theta = 16\tan^{2}{\theta\sin^{2}\theta}}$
Hence, the option (c) satisfies