# Prove the identity, $\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}=\cot {{36}^{\circ }}$

Last updated date: 29th Mar 2023

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Answer

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

Complete step-by-step answer:

Use relation $\cot \left( 90-\theta \right)=\tan \theta $ or $\tan \left( 90-\theta \right)=\cot \theta $ for simplifying it further.

We have to prove that,

$\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}=\cot {{36}^{\circ }}..........\left( i \right)$

Let us prove the given relation by simplifying LHS of equation (i) we have,

LHS$=\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}$

Dividing numerator and denominator by $\cos {{9}^{\circ }}$, we get,

LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}$

Now, we can divide $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in numerator and similarly, $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in denominator as well. We get,

LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}+\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}-\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}............(ii)$

We have trigonometric identity,

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }...........(iii)$

Hence, equation (ii) can be simplified as

LHS = $\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}}=\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\left( 1 \right)}$

As we know, $\tan {{45}^{\circ }}=1$ , so above equation can be written as

LHS = $\dfrac{\tan {{45}^{\circ }}+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\tan {{45}^{\circ }}}.............\left( iv \right)$

Now, using trigonometric identity,

$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A+\tan B}$

Hence equation (iv) can be rewritten with the help of the above equation. So, we get

LHS = $\tan \left( 45+9 \right)=\tan {{54}^{\circ }}$

Now, we can use relation $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ to convert the ‘tan’ function to ‘cot’.

Hence, LHS can be given as

LHS = $\tan \left( 90-36 \right)=\cot {{36}^{\circ }}$

Therefore, LHS = RHS = $\cot {{36}^{\circ }}$

Hence proved the given relation

Note: Another approach for the given question would be to use trigonometric identities

$\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$

$\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$

Use relation $\cos \theta =\sin \left( 90-\theta \right)$

So, LHS = $\dfrac{\cos {{9}^{\circ }}+\cos {{81}^{\circ }}}{\cos 9-\cos 81}$

One can go wrong while converting $\dfrac{\tan 45+\tan {{9}^{\circ }}}{1-\tan 9\tan 45}$ to $\tan {{54}^{\circ }}$ .

He or she may confuse between formula tan(A-B) and tan(A+B).

Complete step-by-step answer:

Use relation $\cot \left( 90-\theta \right)=\tan \theta $ or $\tan \left( 90-\theta \right)=\cot \theta $ for simplifying it further.

We have to prove that,

$\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}=\cot {{36}^{\circ }}..........\left( i \right)$

Let us prove the given relation by simplifying LHS of equation (i) we have,

LHS$=\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}$

Dividing numerator and denominator by $\cos {{9}^{\circ }}$, we get,

LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}$

Now, we can divide $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in numerator and similarly, $\cos {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ and $\sin {{9}^{\circ }}$ by $\cos {{9}^{\circ }}$ in denominator as well. We get,

LHS = $\dfrac{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}+\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}{\dfrac{\cos {{9}^{\circ }}}{\cos {{9}^{\circ }}}-\dfrac{\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}}}............(ii)$

We have trigonometric identity,

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }...........(iii)$

Hence, equation (ii) can be simplified as

LHS = $\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}}=\dfrac{1+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\left( 1 \right)}$

As we know, $\tan {{45}^{\circ }}=1$ , so above equation can be written as

LHS = $\dfrac{\tan {{45}^{\circ }}+\tan {{9}^{\circ }}}{1-\tan {{9}^{\circ }}\tan {{45}^{\circ }}}.............\left( iv \right)$

Now, using trigonometric identity,

$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A+\tan B}$

Hence equation (iv) can be rewritten with the help of the above equation. So, we get

LHS = $\tan \left( 45+9 \right)=\tan {{54}^{\circ }}$

Now, we can use relation $\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta $ to convert the ‘tan’ function to ‘cot’.

Hence, LHS can be given as

LHS = $\tan \left( 90-36 \right)=\cot {{36}^{\circ }}$

Therefore, LHS = RHS = $\cot {{36}^{\circ }}$

Hence proved the given relation

Note: Another approach for the given question would be to use trigonometric identities

$\cos C+\cos D=2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2}$

$\cos C-\cos D=-2\sin \dfrac{C+D}{2}\sin \dfrac{C-D}{2}$

Use relation $\cos \theta =\sin \left( 90-\theta \right)$

So, LHS = $\dfrac{\cos {{9}^{\circ }}+\cos {{81}^{\circ }}}{\cos 9-\cos 81}$

One can go wrong while converting $\dfrac{\tan 45+\tan {{9}^{\circ }}}{1-\tan 9\tan 45}$ to $\tan {{54}^{\circ }}$ .

He or she may confuse between formula tan(A-B) and tan(A+B).

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