Question

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

Answer 1

In the given question, we have been asked to find the value of a trigonometric ratio with a constant angle. We have been given a formula and we have to use it to solve for another angle. This is achieved by just putting the values into the expression, simplifying the function and finding the answer of the given argument of the question.

Formula used:
We are going to use the given formula,
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$

Complete step by step solution:
We have been given the formula,
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
We have to find the value of $\tan 60^\circ$ being given that $\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$.
So, putting $\tan 30^\circ = \dfrac{1}{{\sqrt 3 }}$ in the above formula, we have,
$\tan 60^\circ = \dfrac{{2\tan 30^\circ }}{{1 - {{\tan }^2}30^\circ }}$
Putting the value,
$\tan 60^\circ = \dfrac{{2 \times \dfrac{1}{{\sqrt 3 }}}}{{1 - {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}}} = \dfrac{{\dfrac{2}{{\sqrt 3 }}}}{{1 - \dfrac{1}{3}}} = \dfrac{{\dfrac{2}{{\sqrt 3 }}}}{{\dfrac{2}{3}}}$
Taking the denominator to the numerator,
$\tan 60^\circ = \dfrac{2}{{\sqrt 3 }} \times \dfrac{3}{2} = \sqrt 3$
Hence, the value of $\tan 60^\circ$ is $\sqrt 3$.

Note: In the given question, we were given a formula and using that, we had to find the value of a given expression. This is really simple – just put in the values into the expression, simplify the expression and find the answer to the given question. We have to give care if the denominator is irrational, if it is then we have to rationalize. When we are rationalizing the denominator as it is the only place where we could make an error.