How do you find the exact functional value $\tan \left( {\dfrac{{ - 5\pi }}{{12}}} \right)$using cosine sum or difference identity?

Hint: In order to determine the exact value of $\tan \left( {\dfrac{{ - 5\pi }}{{12}}} \right)$, first find out the value of$\tan \left( {\dfrac{{5\pi }}{{12}}} \right)$by splitting the angle $\dfrac{{5\pi }}{{12}}$as $\dfrac{\pi }{6} + \dfrac{\pi }{4}$ . Now apply the sum of angle formula of tangent $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$ to find the value of$\tan \left( {\dfrac{{5\pi }}{{12}}} \right)$. Now using the property of tangent as $\tan \left( { - \theta } \right) = - \tan \left( \theta \right)$,you’ll get your required result.

Formula Used:
${\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB$
$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$(a - b)(a + b) = {a^2} - {b^2}$

Complete step-by-step solution:
In order the find the exact value of $\tan \left( {\dfrac{{ - 5\pi }}{{12}}} \right)$, We will be first finding the value of $\tan \left( {\dfrac{{5\pi }}{{12}}} \right)$and to do so we have to find the two angles whose either Sum or difference is $\dfrac{{5\pi }}{{12}}$
We only know the exact value of tangent at angles $0,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3},\dfrac{\pi }{2}$.
Now have to find such combination of two angles from the above angles, so that the sum or difference is a $\dfrac{{5\pi }}{{12}}$.
We can write $\dfrac{{5\pi }}{{12}}$as $\dfrac{\pi }{6} + \dfrac{\pi }{4}$
SO we get
$ \Rightarrow \tan \left( {\dfrac{{5\pi }}{{12}}} \right) = \tan \left( {\dfrac{\pi }{6} + \dfrac{\pi }{4}} \right)$
Now Using sum of angle formula for tangent as $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$,by considering A as $\dfrac{\pi }{6}$and B as $\dfrac{\pi }{4}$.
We get,
$\Rightarrow \tan \left( {\dfrac{\pi }{6} + \dfrac{\pi }{4}} \right) = \dfrac{{\tan \left( {\dfrac{\pi }{6}} \right) + \tan \left( {\dfrac{\pi }{4}} \right)}}{{1 - \tan \left( {\dfrac{\pi }{6}} \right)\tan \left( {\dfrac{\pi }{4}} \right)}}$
Since, As we know $\tan \left( {\dfrac{\pi }{6}} \right) = \dfrac{{\sqrt 3 }}{3}$and $\tan \left( {\dfrac{\pi }{4}} \right) = 1$.Putting these values in the above equation we get
$\Rightarrow \dfrac{{1 + \dfrac{{\sqrt 3 }}{3}}}{{1 - \left( {\dfrac{{\sqrt 3 }}{3}} \right)\left( 1 \right)}}$
To simplify the above, multiply and divide the above equation with $1 + \dfrac{{\sqrt 3 }}{3}$, we get
$
  \Rightarrow \dfrac{{1 + \dfrac{{\sqrt 3 }}{3}}}{{1 - \dfrac{{\sqrt 3 }}{3}}} \times \dfrac{{1 + \dfrac{{\sqrt 3 }}{3}}}{{1 + \dfrac{{\sqrt 3 }}{3}}} \\
  \Rightarrow \dfrac{{{{\left( {1 + \dfrac{{\sqrt 3 }}{3}} \right)}^2}}}{{\left( {1 - \dfrac{{\sqrt 3 }}{3}} \right)\left( {1 + \dfrac{{\sqrt 3 }}{3}} \right)}} \\
 $
Now apply the formula of ${\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB$in the numerator to expand and rewrite the denominator using identity $\left( {A - B} \right)\left( {A + B} \right) = {A^2} - {B^2}$. We get our equation as
\[
  \Rightarrow \dfrac{{1 + \dfrac{{2\sqrt 3 }}{3} + \dfrac{1}{3}}}{{1 - \dfrac{1}{3}}} \\
   \Rightarrow \dfrac{{\dfrac{{2\sqrt 3 }}{3} + \dfrac{4}{3}}}{{\dfrac{2}{3}}} \\
   \Rightarrow \dfrac{3}{2}\times \left( {\dfrac{{2\sqrt 3 }}{3} + \dfrac{4}{3}} \right) \\
   \Rightarrow \dfrac{6 \sqrt{3}}{6} + \dfrac{12}{6}
 \]
Simplifying the above further we get
$\tan \left( {\dfrac{{5\pi }}{{12}}} \right) = 2 + \sqrt 3 $
Hence, we have obtained the value of $\tan \left( {\dfrac{{5\pi }}{{12}}} \right) = 2 + \sqrt 3 $
Using the property of tangent that $\tan \left( { - \theta } \right) = - \tan \left( \theta \right)$, we can write $\tan \left( {\dfrac{{ - 5\pi }}{{12}}} \right)$as
$\Rightarrow \tan \left( {\dfrac{{ - 5\pi }}{{12}}} \right) = - \left( {2 + \sqrt 3 } \right) = - 2 - \sqrt 3 $

Therefore, the exact functional value of $\tan \left( {\dfrac{{ - 5\pi }}{{12}}} \right)$is equal to $ - 2 - \sqrt 3 $.

Additional Information:
1. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
2. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore,$\sin \theta $ and $\tan \theta $ and their reciprocals,$\csc \theta $ and $\cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.

Note:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
3. Tangent is always positive in the 1st and 3rd quadrant and negative in 2nd and 4th quadrant.