Question

Using the formula, $tan2A = \dfrac{{2tanA}}{{1 - ta{n^2}A}}$ , Find the value of $tan60^\circ$ , it is being given that $tan30^\circ = \dfrac{1}{{\sqrt 3 }}$

Answer 1

It is a question of trigonometric identity. Use the given formula to get the value of $tan60^\circ$ . We put the value of $tan30^\circ$ as $tanA$ and then follow the steps. Using this formula we may get different values of $tan$ at different angles. The value must be the same as that of the trigonometric table studied earlier.

Complete step by step solution:
We are given with the formula of $tan2A$ and we have to find the value of $tan60$ using this formula $tan2A = \dfrac{{2tanA}}{{1 - ta{n^2}A}}$
Provided that $tan30^\circ = \dfrac{1}{{\sqrt 3 }}$
As we have $60^\circ = 2 \times 30^\circ$
So it implies that if $A = 30^\circ$ then clearly $2A = 2 \times 30^\circ = 60^\circ$
So $tanA = tan30^\circ$ and $tanB = tan60^\circ$
Now we will substitute the angle measure in the given formula to get the result.
On substituting we get,
$tan60^\circ = \dfrac{{2tan30^\circ }}{{1 - ta{n^2}30^\circ }}$
As we have $tan30^\circ = \dfrac{1}{{\sqrt 3 }}$
Then, putting this value in the formula we get,
$\Rightarrow tan60^\circ = \dfrac{{2\left( {\dfrac{1}{{\sqrt 3 }}} \right)}}{{1 - {{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2}}}$
Simplifying by Squaring in denominator we get,
$\Rightarrow tan60^\circ = \dfrac{{2\left( {\dfrac{1}{{\sqrt 3 }}} \right)}}{{1 - \dfrac{1}{3}}}$
Now taking LCM in denominator and simplifying
$\Rightarrow tan60^\circ = \dfrac{{2\left( {\dfrac{1}{{\sqrt 3 }}} \right)}}{{\dfrac{{3 - 1}}{3}}} = \dfrac{{2\left( {\dfrac{1}{{\sqrt 3 }}} \right)}}{{\left( {\dfrac{2}{3}} \right)}}$
Now as it is division of fractions so we have to multiply the numerator of the whole fraction with the reciprocal of denominator of the fraction
It will be then,
$\Rightarrow tan60^\circ = \dfrac{2}{{\sqrt 3 }} \times \dfrac{3}{2}$
On simple multiplication and division (cancelling out)
$\Rightarrow tan60^\circ = \sqrt 3$
Hence by using the above formula the value of $tan60^\circ$ is $\sqrt 3$
So, the correct answer is “$\sqrt 3$”.

Note : This formula works for the angle measures in degree as well as in radians. This is the derived formula of $tan\left( {A + B} \right) = \dfrac{{tanA + tanB}}{{1 - tanA \times tanB}}$ Here $B$ is replaced by $A$ itself and hence we get the formula for $tan2A$ . The range of tangent function is $\mathbb{R}$ that is the set of real numbers. We can similarly obtain the values of other angles as well, even multiples of $30^\circ$ are obtained by using this formula itself with the given value of $tan30^\circ$ .