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 \\
$

Answer 1

Hint: - Use $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Given equation is
$\dfrac{{2\tan 30^\circ }}{{1 - {{\tan }^2}30^\circ }}$
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 $
\[ \Rightarrow \dfrac{{2\sin 30^\circ \cos 30^\circ }}{{{{\cos }^2}30^\circ - {{\sin }^2}30^\circ }} = \dfrac{{\sin 60^\circ }}{{\cos 60^\circ }} = \tan 60^\circ \]
So, this is the required answer.
Hence, option (c) is correct.

Note: - Whenever we face such types of problems, always remember the trigonometric identities which are written above then simplify the given problem statement using these identities we will get the required answer.