Question

Find the value of \[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}}\] is equal to
A. \[\tan {26^ \circ }\]
B. \[\tan {81^ \circ }\]
C. \[\tan {51^ \circ }\]
D. \[\tan {54^ \circ }\]
E. \[\tan {46^ \circ }\]

Answer 1

In the given question, we are asked to find the value of \[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}}\]. For that, we divide the numerator and denominator by \[\cos {9^ \circ }\] then simplify it to get the desired result.

Formula used:
We have been using the following formulas:
1. \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
2. \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\]

Complete step-by-step solution:
We are given that \[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}}\]
We need to find the value of the given function.
Now we divide the numerator and denominator by \[\cos {9^ \circ }\], we will get
\[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ }}}}}{{\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ }}}}}...\left( 1 \right)\]
Now we can divide \[\cos {9^ \circ }\]by \[\cos {9^ \circ }\]and \[\sin {9^ \circ }\]by \[\sin {9^ \circ }\] in the numerator and similarly, \[\cos {9^ \circ }\]by \[\cos {9^ \circ }\] and \[\sin {9^ \circ }\] by \[\sin {9^ \circ }\] in the denominator, we get
\[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{\dfrac{{\cos {9^ \circ }}}{{\cos {9^ \circ }}} + \dfrac{{\sin {9^ \circ }}}{{\cos {9^ \circ }}}}}{{\dfrac{{\cos {9^ \circ }}}{{\cos {9^ \circ }}} - \dfrac{{\sin {9^ \circ }}}{{\cos {9^ \circ }}}}}...\left( 2 \right)\]
Now we know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Now we apply above formula in equation and cancel the like terms, we will get
\[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{1 + \tan {9^ \circ }}}{{1 - \tan {9^ \circ }\left( 1 \right)}}...\left( 3 \right)\]
Now we know that \[\tan {45^ \circ } = 1\]
Now we substitute this value in our equation, we will get
\[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{\tan {{45}^ \circ } + \tan {9^ \circ }}}{{1 - \tan {9^ \circ }\tan {{45}^ \circ }}}...\left( 4 \right)\]
Now we use trigonometric identity \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] in equation, we will get
\[
  \dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \tan \left( {{{45}^ \circ } + {9^ \circ }} \right) \\
   = \tan {54^ \circ }
 \]
Therefore, the value of \[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}}\] is equal to \[\tan {54^ \circ }\]
Hence, option (D) is correct option

Note: Students also take approach the following formula:
Given that \[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}}...\left( 1 \right)\]
Now we know that \[\cos \left( \theta \right) = \sin \left( {90 - \theta } \right)\]
Now we apply this formula in equation (1), we get
\[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{\sin \left( {{{90}^ \circ } - {9^ \circ }} \right) + \sin {9^ \circ }}}{{\sin \left( {{{90}^ \circ } - {9^ \circ }} \right) - \sin {9^ \circ }}}\]
By simplifying, we get
\[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{\sin {{81}^ \circ } + \sin {9^ \circ }}}{{\sin {{81}^ \circ } - \sin {9^ \circ }}}\]
Now we know that
\[
  \sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right) \\
  \sin A - \sin B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)
 \]
By applying the above formula in our equation and then simplifying it, we get
\[
  \dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{2\sin \left( {\dfrac{{{{81}^ \circ } + {9^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{81}^ \circ } - {9^ \circ }}}{2}} \right)}}{{2\cos \left( {\dfrac{{{{81}^ \circ } + {9^ \circ }}}{2}} \right)\sin \left( {\dfrac{{{{81}^ \circ } - {9^ \circ }}}{2}} \right)}} \\
   = \dfrac{{2\sin \left( {\dfrac{{{{90}^ \circ }}}{2}} \right)\cos \left( {\dfrac{{{{72}^ \circ }}}{2}} \right)}}{{2\cos \left( {\dfrac{{{{90}^ \circ }}}{2}} \right)\sin \left( {\dfrac{{{{72}^ \circ }}}{2}} \right)}} \\
   = \dfrac{{2\sin {{45}^ \circ }\cos {{36}^ \circ }}}{{2\cos {{45}^ \circ }\sin {{36}^ \circ }}} \\
   = \dfrac{{\sin {{45}^ \circ }\cos {{36}^ \circ }}}{{\cos {{45}^ \circ }\sin {{36}^ \circ }}}
 \]
Now we know that \[\sin {45^ \circ } = \cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\].
Therefore, our equation becomes
 \[
  \dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \dfrac{{\cos {{36}^ \circ }}}{{\sin {{36}^ \circ }}} \]
We know that \[\cot \theta={\dfrac{\cos \theta}{\sin \theta}}\]
\[= \cot {36^ \circ } \]
We know that \[\cot \theta = \tan \left( {{{90}^ \circ } - \theta } \right)\]
Therefore, our equation becomes
\[
  \dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}} = \tan \left( {{{90}^ \circ } - {{36}^ \circ }} \right) \\
   = \tan {54^ \circ }
 \]
Therefore, the value of \[\dfrac{{\cos {9^ \circ } + \sin {9^ \circ }}}{{\cos {9^ \circ } - \sin {9^ \circ }}}\] is equal to \[\tan {54^ \circ }\]