Question

Prove that \[\dfrac{{\cos {{17}^0} + \sin {{17}^0}}}{{\cos {{17}^0} - \sin {{17}^0}}} = \tan {62^0}\] .

Answer 1

Hint: Divide in numerator and denominator by \[\cos {17^0}\] ,we will get the expression in terms of \[\tan {17^0}\] .Since we have to prove the final answer in terms of tan only so we will manipulate the expression using the identity \[\tan ({45^0} + \theta ) = \dfrac{{1 + \tan \theta }}{{1 - \tan \theta }}\] to get the required answer.

Formula Used
 1. \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
2.\[\]\[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]

Complete step by step solution
Given – LHS=\[\dfrac{{\cos {{17}^0} + \sin {{17}^0}}}{{\cos {{17}^0} - \sin {{17}^0}}}\]
Dividing by \[\cos {17^0}\] in numerator and denominator
               LHS=\[\dfrac{{\dfrac{{\cos {{17}^0}}}{{\cos {{17}^0}}} + \dfrac{{\sin {{17}^0}}}{{\cos {{17}^0}}}}}{{\dfrac{{\cos 17}}{{\cos {{17}^0}}} - \dfrac{{\sin {{17}^0}}}{{\cos {{17}^0}}}}}\]
 We know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] ,using these formulae in above expression
                  LHS= \[\dfrac{{1 + \tan {{17}^0}}}{{1 - \tan {{17}^0}}}\]
We know that \[\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]
Put \[A = {45^0}\] and \[B = {17^0}\] in the above formula
                   \[\tan ({45^0} + {17^0}) = \dfrac{{\tan {{45}^0} + \tan {{17}^0}}}{{1 - \tan {{45}^0}\tan {{17}^0}}}\]
But \[\tan {45^0} = 1\]
∴ \[\tan ({45^0} + {17^0}) = \dfrac{{1 + \tan {{17}^0}}}{{1 - \tan {{17}^0}}}\]
But \[\dfrac{{1 + \tan {{17}^0}}}{{1 - \tan {{17}^0}}}\]=LHS
\[ \Rightarrow \tan ({45^0} + {17^0})\] =LHS
 \[ \Rightarrow \]LHS=\[\tan {62^0}\]
But \[\tan {62^0} = \] RHS
Thus LHS=RHS
Hence proved that \[\dfrac{{\cos {{17}^0} + \sin {{17}^0}}}{{\cos {{17}^0} - \sin {{17}^0}}} = \tan {62^0}\].

Note: This question can be solved by another method as
                  LHS=\[\dfrac{{\cos {{17}^0} + \sin {{17}^0}}}{{\cos {{17}^0} - \sin {{17}^0}}}\]
Multiplying by \[\cos {17^0} + \sin {17^0}\] in numerator and denominator
       LHS=\[\dfrac{{{{\left( {\cos {{17}^0} + \sin {{17}^0}} \right)}^2}}}{{{{\cos }^2}{{17}^0} - {{\sin }^2}{{17}^2}}}\]
       LHS=\[\dfrac{{{{\cos }^2}{{17}^0} + {{\sin }^2}{{17}^0} + 2\sin {{17}^0}\cos {{17}^0}}}{{\cos {{34}^0}}}\]
         \[\left( \begin{array}{l}{\cos ^2}A + {\sin ^2}A = 1\\{\cos ^2}A - {\sin ^2}A = \cos 2A\end{array} \right)\]
        LHS=\[\dfrac{{1 + \sin {{34}^0}}}{{\cos {{34}^0}}}\]
But \[\sin {34^0} = \cos {56^0}\] so ,
      LHS= \[\dfrac{{1 + \cos {{56}^0}}}{{\cos {{34}^0}}}\]
But \[1 + \cos 2A = {\cos ^2}A\]
∴ LHS= \[\dfrac{{2{{\cos }^2}{{28}^0}}}{{\cos {{34}^0}}}\]
As \[\cos {34^0} = \sin {56^0}\]
LHS= \[\dfrac{{2{{\cos }^2}{{28}^0}}}{{\sin {{56}^0}}}\]
Using \[\sin 2A = 2\sin A\cos A\]
LHS= \[\dfrac{{2{{\cos }^2}{{28}^0}}}{{2\sin {{28}^0}\cos {{28}^0}}}\]
Simplifying the expression
LHS= \[\dfrac{{\cos {{28}^0}}}{{\sin {{28}^0}}}\]
LHS= \[\cot {28^0}\]
But \[\cot \theta = \tan (90 - \theta )\]
Therefore
LHS= \[\tan {62^0}\]
Hence proved LHS=RHS.