Question

The value of $\dfrac{2\tan {{30}^{\circ }}}{1-{{\tan }^{2}}{{30}^{\circ }}}$ is equal to:
(a) $\cos {{60}^{\circ }}$
(b) $\sin {{60}^{\circ }}$
(c) $\tan {{60}^{\circ }}$
(d) $\sin {{30}^{\circ }}$

Answer 1

Hint:In the given expression, first of all find the value of $\tan {{30}^{0}}$.From the trigonometric functions, we know the value of $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ then substitute this value in the given expression and simplify.

Complete step-by-step answer:
The expression given in the question that we have to find value of is:
$\dfrac{2\tan {{30}^{\circ }}}{1-{{\tan }^{2}}{{30}^{\circ }}}$
The value of tan 30° is determined from the table of trigonometric functions which says $\tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}}$ .
Plugging this value of $\tan {{30}^{0}}$ in the above expression we get,
$\begin{align}
  & \dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}} \\
 & =\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \\
\end{align}$
$=\sqrt{3}$
From the above calculation, the value of the given expression is $\sqrt{3}$ .
Now, we are going to compare the result of the options with the result of the given expression.
(a) $\cos {{60}^{\circ }}=\dfrac{1}{2}$
The answer of the above option is not matched with the answer of the given expression.
(b) $\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}$
The answer of the above option is not matched with the answer of the given expression.
(c) $\tan {{60}^{\circ }}=\sqrt{3}$
The answer of the above option is matched with the answer of the given expression.
(d) $\sin {{30}^{\circ }}=\dfrac{1}{2}$
The answer of the above option is not matched with the answer of the given expression.
From the above discussion, we have found that the answer of the option (c) is equal to the given expression.
Hence, the correct option is (c).

Note: The alternate way of solving the above problem is discussed below.
The expression given in the above question is:
$\dfrac{2\tan {{30}^{\circ }}}{1-{{\tan }^{2}}{{30}^{\circ }}}$
If you can recall the identity of $\tan 2\theta $ , the above expression is the expansion of $\tan 2\theta $ .
$\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$
If we compare the above expression with the expression given in the question, we can see that the value of $\theta ={{30}^{0}}$ .So, the given expression is equal to:
$\dfrac{2\tan {{30}^{\circ }}}{1-{{\tan }^{2}}{{30}^{\circ }}}=\tan {{60}^{\circ }}$
   From the above calculation, we have found that the simplification of the given expression yields to $\tan {{60}^{0}}$ .
Hence, the correct option is (c) which is equal to $\tan {{60}^{0}}$ .