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Question: Prove the identity, \(\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^...

Prove the identity, cos⁡9∘+sin⁡9∘cos⁡9∘−sin⁡9∘=cot⁡36∘\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}=\cot {{36}^{\circ }}

Explanation

Solution

Hint: Divide numerator and denominator by cos⁡9∘\cos {{9}^{\circ }} or sin⁡9∘\sin {{9}^{\circ }} to get a trigonometric identity.

Complete step-by-step answer:
Use relation cot⁡(90−θ)=tan⁡θ\cot \left( 90-\theta \right)=\tan \theta or tan⁡(90−θ)=cot⁡θ\tan \left( 90-\theta \right)=\cot \theta for simplifying it further.

We have to prove that,
cos⁡9∘+sin⁡9∘cos⁡9∘−sin⁡9∘=cot⁡36∘..........(i)\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}=\cot {{36}^{\circ }}..........\left( i \right)
Let us prove the given relation by simplifying LHS of equation (i) we have,
LHS=cos⁡9∘+sin⁡9∘cos⁡9∘−sin⁡9∘=\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}
Dividing numerator and denominator by cos⁡9∘\cos {{9}^{\circ }}, we get,
LHS = cos⁡9∘+sin⁡9∘cos⁡9∘cos⁡9∘−sin⁡9∘cos⁡9∘\dfrac{\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}
Now, we can divide cos⁡9∘\cos {{9}^{\circ }} by cos⁡9∘\cos {{9}^{\circ }} and sin⁡9∘\sin {{9}^{\circ }} by cos⁡9∘\cos {{9}^{\circ }} in numerator and similarly, cos⁡9∘\cos {{9}^{\circ }} by cos⁡9∘\cos {{9}^{\circ }} and sin⁡9∘\sin {{9}^{\circ }} by cos⁡9∘\cos {{9}^{\circ }} in denominator as well. We get,
LHS = cos⁡9∘cos⁡9∘+sin⁡9∘cos⁡9∘cos⁡9∘cos⁡9∘−sin⁡9∘cos⁡9∘............(ii)\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 }}}}............(ii)
We have trigonometric identity,
tan⁡θ=sin⁡θcos⁡θ...........(iii)\tan \theta =\dfrac{\sin \theta }{\cos \theta }...........(iii)
Hence, equation (ii) can be simplified as
LHS = 1+tan⁡9∘1−tan⁡9∘=1+tan⁡9∘1−tan⁡9∘(1)\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}}=\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\left( 1 \right)}
As we know, tan⁡45∘=1\tan {{45}^{\circ }}=1 , so above equation can be written as
LHS = tan⁡45∘+tan⁡9∘1−tan⁡9∘tan⁡45∘.............(iv)\dfrac{\tan {{45}^{\circ }}+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\tan {{45}^{\circ }}}.............\left( iv \right)
Now, using trigonometric identity,
tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A+tan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A+\tan B}
Hence equation (iv) can be rewritten with the help of the above equation. So, we get
LHS = tan⁡(45+9)=tan⁡54∘\tan \left( 45+9 \right)=\tan {{54}^{\circ }}
Now, we can use relation tan⁡(90∘−θ)=cot⁡θ\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta to convert the ‘tan’ function to ‘cot’.
Hence, LHS can be given as
LHS = tan⁡(90−36)=cot⁡36∘\tan \left( 90-36 \right)=\cot {{36}^{\circ }}
Therefore, LHS = RHS = cot⁡36∘\cot {{36}^{\circ }}
Hence proved the given relation

Note: Another approach for the given question would be to use trigonometric identities
cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D2\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}
cos⁡C−cos⁡D=−2sin⁡C+D2sin⁡C−D2\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}
Use relation cos⁡θ=sin⁡(90−θ)\cos \theta =\sin \left( 90-\theta \right)
So, LHS = cos⁡9∘+cos⁡81∘cos⁡9−cos⁡81\dfrac{\cos {{9}^{\circ }}+\cos {{81}^{\circ }}}{\cos 9-\cos 81}
One can go wrong while converting tan⁡45+tan⁡9∘1−tan⁡9tan⁡45\dfrac{\tan 45+\tan {{9}^{\circ }}}{1-\tan 9\tan 45} to tan⁡54∘\tan {{54}^{\circ }} .
He or she may confuse between formula tan(A-B) and tan(A+B).