Question

Find the value of $\tan \dfrac{{13\pi }}{{12}}$.

Answer 1

We have to find the value of the given trigonometric function $\tan \dfrac{{13\pi }}{{12}}$ . We solve this question using the concept of values of trigonometric function in various quadrants . The formula for the difference of the two angles of a tangent function . We should also have the knowledge of various values of trigonometric functions for different values of angles . First we will simplify the value of the angle of the given trigonometric function using the value of the trigonometric values in different quadrants , then we will split the angle of the given trigonometric function such that we know the exact values for that angle . Then applying the formula of the difference of the tangent function and putting the values we will get the value for the given trigonometric function.

Complete step by step answer:
Given: The value of $\tan \dfrac{{13\pi }}{{12}}$. Now , we can write the given expression as : $\tan \dfrac{{13\pi }}{{12}} = \tan \left( {\pi + \dfrac{\pi }{{12}}} \right)$
Using the quadrant rules we can say that the given angle of the trigonometric function lies in the third quadrant and we know that the tan function is positive in the third quadrant so we can also write the given expression as :
$\tan \dfrac{{13\pi }}{{12}} = \tan \dfrac{\pi }{{12}}$
Now , we have to find the value of $\tan \dfrac{\pi }{{12}}$ .
We can also express $\tan \dfrac{\pi }{{12}}$ as :
$\tan \dfrac{\pi }{{12}} = \tan \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$

Now , we know that the formula for difference of two angles of tangent function is given as :
$\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a \times \tan b}}$
Using the formula , we get
$\tan \dfrac{\pi }{{12}} = \dfrac{{\tan \dfrac{\pi }{3} - \tan \dfrac{\pi }{4}}}{{1 + \tan \dfrac{\pi }{3} \times \tan \dfrac{\pi }{4}}}$
We also know the value of tangent function for different angles are given as :
$\tan \dfrac{\pi }{3} = \sqrt 3$ and $\tan \dfrac{\pi }{4} = 1$
Putting these values in the formula , we get
$\tan \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 \times 1}}$
$\Rightarrow \tan \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }}$

Now , we will rationalize the value of $\tan \dfrac{\pi }{3} = \sqrt 3$ by multiplying both
numerator and denominator by $\sqrt 3 - 1$ .On rationalizing , we get
$\tan \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}$
On simplifying using the formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
$\tan \dfrac{\pi }{{12}} = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}}$
On further solving , we get
$\tan \dfrac{\pi }{{12}} = \dfrac{{\left( {3 + 1 - 2\sqrt 3 } \right)}}{{3 - 1}}$
$\Rightarrow \tan \dfrac{\pi }{{12}} = \dfrac{{\left( {4 - 2\sqrt 3 } \right)}}{2}$
Cancelling the terms , we get
$\therefore \tan \dfrac{\pi }{{12}} = 2 - \sqrt 3$

Hence,the value of $\tan \dfrac{{13\pi }}{{12}}$ is $2 - \sqrt 3$.

Note: While using the formula of the difference of angles of the tangent we should take care of the signs while writing them. A slight change in any of the signs changes the whole answer. The values should be solved carefully and while using all the values should be taken care of. The formula for sum of two angles of tangent function is given as: $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \times \tan b}}$