Question

# The value of $\dfrac{{2\tan 30^\circ }}{{1 - {{\tan }^2}30^\circ }}$ is$a.{\text{ }}\cos 60^\circ \\ b.{\text{ sin}}60^\circ \\ c.{\text{ tan}}60^\circ \\ d.{\text{ sin3}}0^\circ \\$

Hint: - Use $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\dfrac{{2\tan 30^\circ }}{{1 - {{\tan }^2}30^\circ }}$
Now substitute $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\Rightarrow \dfrac{{2\dfrac{{\sin 30^\circ }}{{\cos 30^\circ }}}}{{1 - {{\left( {\dfrac{{\sin 30^\circ }}{{\cos 30^\circ }}} \right)}^2}}} = \dfrac{{2\dfrac{{\sin 30^\circ }}{{\cos 30^\circ }}}}{{\dfrac{{{{\cos }^2}30^\circ - {{\sin }^2}30^\circ }}{{{{\cos }^2}30^\circ }}}} = \dfrac{{2\sin 30^\circ }}{{{{\cos }^2}30^\circ - {{\sin }^2}30^\circ }} \times \dfrac{{{{\cos }^2}30^\circ }}{{\cos 30^\circ }} \\ \Rightarrow \dfrac{{2\sin 30^\circ \cos 30^\circ }}{{{{\cos }^2}30^\circ - {{\sin }^2}30^\circ }} \\$
As we know ${\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta ,{\text{ 2sin}}\theta {\text{cos}}\theta = \sin 2\theta$
$\Rightarrow \dfrac{{2\sin 30^\circ \cos 30^\circ }}{{{{\cos }^2}30^\circ - {{\sin }^2}30^\circ }} = \dfrac{{\sin 60^\circ }}{{\cos 60^\circ }} = \tan 60^\circ$