Prove that \[\dfrac{{\cos 8^\circ - \sin 8^\circ }}{{\cos 8^\circ + \sin 8^\circ }} = \tan 37^\circ \]

Hint:
Here, we are required to prove the given equation. Thus, we will divide the left hand side of the equation by \[\cos 8^\circ \] and simplify it further to get the equation in form of tangent function. Then using the suitable trigonometric identity we will simplify the equation further so that the expression on the left hand side of the given equation is equal to the right hand side.

Formula Used:
We will use the following formulas:
1) \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
2) \[\dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}} = \tan \left( {a - b} \right)\]

Complete step by step solution:
We will first consider the left hand side of the given equation.
LHS \[ = \dfrac{{\cos 8^\circ - \sin 8^\circ }}{{\cos 8^\circ + \sin 8^\circ }}\]
Now, dividing both the numerator as well as denominator by \[\cos 8^\circ \], we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{\dfrac{{\cos 8^\circ }}{{\cos 8^\circ }} - \dfrac{{\sin 8^\circ }}{{\cos 8^\circ }}}}{{\dfrac{{\cos 8^\circ }}{{\cos 8^\circ }} + \dfrac{{\sin 8^\circ }}{{\cos 8^\circ }}}}\]
Now using the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\], we get
\[ \Rightarrow \] LHS \[ = \dfrac{{1 - \tan 8^\circ }}{{1 + \tan 8^\circ }}\]
Now, we know that \[\tan 45^\circ = 1\]
Using this we can write above equation as:
\[ \Rightarrow \] LHS \[ = \dfrac{{\tan 45^\circ - \tan 8^\circ }}{{1 + \tan 45^\circ \tan 8^\circ }}\]
Here, using the formula, \[\dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}} = \tan \left( {a - b} \right)\], we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{\tan 45^\circ - \tan 8^\circ }}{{1 + \tan 45^\circ \tan 8^\circ }} = \tan \left( {45^\circ - 8^\circ } \right)\]
\[ \Rightarrow \] LHS \[ = \tan 37^\circ = \] RHS
Hence,
LHS \[ = \] RHS
\[\dfrac{{\cos 8^\circ - \sin 8^\circ }}{{\cos 8^\circ + \sin 8^\circ }} = \tan 37^\circ \]
Hence, proved

Additional Information:
Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life.

Note:
An alternate way to solve this question is:
We have,
LHS \[ = \dfrac{{\cos 8^\circ - \sin 8^\circ }}{{\cos 8^\circ + \sin 8^\circ }}\]
Rewriting this expression, we get
\[ \Rightarrow \] LHS \[ = \dfrac{{\cos \left( {90^\circ - 82^\circ } \right) - \sin 8^\circ }}{{\cos \left( {90^\circ - 82^\circ } \right) + \sin 8^\circ }}\]
Hence, using the formula, \[\cos \left( {90^\circ - \theta } \right) = \sin \theta \], we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{\sin 82^\circ - \sin 8^\circ }}{{\sin 82^\circ + \sin 8^\circ }}\]
Using the formulas, \[\sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)\] and \[\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)\] in the numerator and the denominator respectively, we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{2\cos \left( {\dfrac{{82^\circ + 8^\circ }}{2}} \right)\sin \left( {\dfrac{{82^\circ - 8^\circ }}{2}} \right)}}{{2\sin \left( {\dfrac{{82^\circ + 8^\circ }}{2}} \right)\cos \left( {\dfrac{{82^\circ - 8^\circ }}{2}} \right)}}\]
Simplifying the expression, we get
\[ \Rightarrow \] LHS\[ = \dfrac{{2\cos 45^\circ \sin 37^\circ }}{{2\sin 45^\circ \cos 37^\circ }}\]
Substituting \[\sin 45^\circ = \cos 45^\circ = \dfrac{1}{{\sqrt 2 }}\] in the above equation, we get
\[ \Rightarrow \] LHS \[ = \dfrac{{2\left( {\dfrac{1}{{\sqrt 2 }}} \right)\sin 37^\circ }}{{2\left( {\dfrac{1}{{\sqrt 2 }}} \right)\cos 37^\circ }}\]
Cancelling out the same terms from the numerator and denominator, we get,
\[ \Rightarrow \] LHS \[ = \dfrac{{\sin 37^\circ }}{{\cos 37^\circ }}\]
Now using the \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\], we get
\[ \Rightarrow \] LHS \[ = \dfrac{{\sin 37^\circ }}{{\cos 37^\circ }} = \tan 37^\circ = \] RHS
Therefore,

\[\dfrac{{\cos 8^\circ - \sin 8^\circ }}{{\cos 8^\circ + \sin 8^\circ }} = \tan 37^\circ \]
Hence, proved.