Question

How do you find the exact value of $\tan \left( {\dfrac{{7\pi }}{{12}}} \right)$?

Answer 1

Hint: We will first write the angle $\dfrac{{7\pi }}{{12}}$ in terms of $\pi - \theta $, then after using the identity \[\tan \left( {\pi - \theta } \right) = - \tan \theta \], we will use the identity $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$.

Complete step by step solution:
We are given that we are required to find the exact value of $\tan \left( {\dfrac{{7\pi }}{{12}}} \right)$.
We can write this as: $\tan \left( {\pi - \dfrac{{5\pi }}{{12}}} \right)$.
Now, we know that we have an identity given by the following expression:-
\[ \Rightarrow \tan \left( {\pi - \theta } \right) = - \tan \theta \]
Therefore, we have: $\tan \left( {\pi - \dfrac{{5\pi }}{{12}}} \right) = - \tan \left( {\dfrac{{5\pi }}{{12}}} \right)$
Now, this can be written as: $\tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \tan \left( {\dfrac{{3\pi }}{{12}} + \dfrac{{2\pi }}{{12}}} \right) = - \tan \left( {\dfrac{\pi }{4} + \dfrac{\pi }{6}} \right)$
Now, we know that we have an identity given by the following expression:-
$ \Rightarrow \tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Replacing A by $\dfrac{\pi }{4}$ and B by $\dfrac{\pi }{6}$ in the above mentioned formula, we will then obtain the following equation with us:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \tan \left( {\dfrac{\pi }{4} + \dfrac{\pi }{6}} \right) = - \left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan \left( {\dfrac{\pi }{6}} \right)}}{{1 - \tan \left( {\dfrac{\pi }{4}} \right)\tan \left( {\dfrac{\pi }{6}} \right)}}} \right)$
Putting in the known values, we will then obtain:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \left( {\dfrac{{1 + \dfrac{1}{{\sqrt 3 }}}}{{1 - 1 \times \dfrac{1}{{\sqrt 3 }}}}} \right)$
Simplifying the denominator, we will then obtain the following equation:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \left( {\dfrac{{1 + \dfrac{1}{{\sqrt 3 }}}}{{1 - \dfrac{1}{{\sqrt 3 }}}}} \right)$
Taking the least common multiples of both the numerator and denominators in the right hand side of the above equation, we will then obtain the following equation:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \left( {\dfrac{{\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}{{\dfrac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}} \right)$
We can write this equation as follows:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \left( {\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right)$
Multiplying and dividing the right hand side of the above equation by $\sqrt 3 + 1$, we will then obtain the following equation:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \left( {\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} \times \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}} \right)$
We will now use the identities ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${a^2} - {b^2} = (a - b)(a + b)$.
Thus, we will then obtain the following equation:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \left( {\dfrac{{3 + 1 + 2\sqrt 3 }}{{3 - 1}}} \right)$
Simplifying the calculations in the numerator and denominator of the right hand side of the above equation, we will then obtain the following equation:-
$ \Rightarrow \tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - \left( {\dfrac{{4 + 2\sqrt 3 }}{2}} \right)$

Hence, the answer is $\tan \left( {\dfrac{{7\pi }}{{12}}} \right) = - 2 - \sqrt 3 $.

Note:-
The students must note that we used the identities ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and ${a^2} - {b^2} = (a - b)(a + b)$ to simplify $ - \left( {\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} \times \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}} \right)$ because, we have:-
\[ \Rightarrow - \left( {\dfrac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} \times \dfrac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}} \right) = - \left[ {\dfrac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}}} \right]\]
Now, we will just replace a by \[\sqrt 3 \] and b by 1 in the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ to obtain the following:-
\[ \Rightarrow {\left( {\sqrt 3 + 1} \right)^2} = 3 + 1 + 2\sqrt 3 \]
Now, we will just replace a by \[\sqrt 3 \] and b by 1 in the formula ${a^2} - {b^2} = (a - b)(a + b)$ to obtain the following:-
\[ \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right) = 3 - 1 = 2\]
Thus, we have the required.