Question

Using the formula, $\tan \left( 2A \right)=\dfrac{2\tan A}{1-{{\tan }^{2}}A}$, find the value of $\tan 60{}^\circ $, it being given that $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Answer 1

Hint: We will be using the concept of trigonometric functions to solve the problem. We will use the fact given to us that $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$ and the formula that $\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}$ to find the value of $\tan 60{}^\circ $.

Complete step-by-step answer:
Now, we have been given to us that $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$ and we have to find the value of $\tan 60{}^\circ $ by using the formula that $\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}$.

Now, let us take the value of $A=30{}^\circ $. Therefore, we have that

$\tan A=\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}..........\left( 1 \right)$

Also, we have been given that,

$\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}...........\left( 2 \right)$

So, we substituted the value of $\tan A$ from (1) in (2).

$\begin{align}

  & \tan 2A=\dfrac{2\times \dfrac{1}{\sqrt{3}}}{1-\dfrac{1}{3}} \\

 & \tan 2A=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}} \\

 & =\dfrac{2\times \sqrt{3}}{2\times \sqrt{3}} \\

 & \tan 2A=\sqrt{3} \\

\end{align}$

Now, we know that the value of $A=30{}^\circ $. Therefore, we have the value of $\tan 60{}^\circ =\sqrt{3}$.


Note: To solve these type of question it is important to note that we have used the data given in the question that $\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}$ to solve the problem.