How do you use the sum or difference identities to find the exact value of \[\tan \left( {\dfrac{\pi }{{12}}} \right)\] ?

Hint: Here in this question to find the exact solution of given trigonometric function by using the formula of tangent difference rule defined as \[tan(A - B) = \dfrac{{tanA - tanB}}{{1 + tanA\,tanB}}\] where A and B are the angles then by using the value of specified angle of trigonometric ratios on simplification, we get the required result.

Complete step-by-step answer:
Angle sum identities and angle difference identities can be used to find the function values of any angles however, the most practical use is to find exact values of an angle that can be written as a sum or difference using the familiar values for the sine, cosine and tangent of the \[\dfrac{\pi }{6}\] , \[\dfrac{\pi }{4}\] , \[\dfrac{\pi }{3}\] and \[\dfrac{\pi }{2}\] angles and their multiples.
sine addition and difference formula can be defined as:
 \[\sin (A + B) = \sin A.\,\cos B + \cos A.\,\sin B\]
 \[\sin (A - B) = \sin A.\,\cos B - \cos A.\,\sin B\]
cosine addition and difference formula can be defined as:
 \[cos\left( {A + B} \right) = cos\,A.cos\,B - sin\,A.sin\,B\]
 \[cos\left( {A - B} \right) = cos\,A.cos\,B + sin\,A.sin\,B\]
Tangent addition and difference formula can be defined as:
 \[tan(A + B) = \dfrac{{tanA + tanB}}{{1 - tanA\,tanB}}\]
 \[tan(A - B) = \dfrac{{tanA - tanB}}{{1 + tanA\,tanB}}\]
Consider the given question
 \[ \Rightarrow \,\,\,\tan \left( {\dfrac{\pi }{{12}}} \right)\]
Can be written as \[\dfrac{\pi }{{12}} = \dfrac{\pi }{3} - \dfrac{\pi }{4}\]
 \[ \Rightarrow \,\,\,\tan \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)\]
Now using the tangent difference formula i.e., \[tan(A - B) = \dfrac{{tanA - tanB}}{{1 + tanA\,tanB}}\]
Where angle A= \[\dfrac{\pi }{3}\] and B= \[\dfrac{\pi }{4}\] .
Therefore, \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \dfrac{{tan\left( {\dfrac{\pi }{3}} \right) - tan\left( {\dfrac{\pi }{4}} \right)}}{{1 + tan\left( {\dfrac{\pi }{3}} \right)tan\left( {\dfrac{\pi }{4}} \right)}}\]
Using the value of specified angle of trigonometric tangent ratio ratios is \[tan\left( {\dfrac{\pi }{3}} \right) = \sqrt 3 \] and \[tan\left( {\dfrac{\pi }{4}} \right) = 1\]
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 \cdot 1}}\]
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }}\]

Multiply and divide \[\sqrt 3 - 1\] , then
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 }} \times \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}\]
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {1 + \sqrt 3 } \right)\left( {\sqrt 3 - 1} \right)}}\]
We know the algebraic formula \[\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)\] and \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\] , then
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\left( {{{\sqrt 3 }^2} - 2 \cdot \sqrt 3 \cdot 1 + {1^2}} \right)}}{{\left( {{{\sqrt 3 }^2} - {1^2}} \right)}}\]
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{3 - 2\sqrt 3 + 1}}{{3 - 1}}\] [ \[\because \,\,{\sqrt 3 ^2} = 3\] ]
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{4 - 2\sqrt 3 }}{2}\]
 \[ \Rightarrow \,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = \dfrac{4}{2} - \dfrac{{2\sqrt 3 }}{2}\]
 \[\therefore \,\,\,\,\,tan\left( {\dfrac{\pi }{{12}}} \right) = 2 - \sqrt 3 \]
So, the correct answer is “$2 - \sqrt 3$”.

Note: In trigonometry, for the trigonometry ratios we have sum and difference formula. For the value of trigonometry ratios, we follow the table of trigonometry ratios for the standard angles. The standard angles either in the form of degree or radian the values will be the same there is no alter or change in the values.