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

Hint: 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 \[tan30^\circ \] .