Prove that: (cos 9° + sin 9°)/(cos 9° − sin 9°) = tan 54°

Question

Prove that:

\[ \frac{\cos 9^\circ+\sin 9^\circ} {\cos 9^\circ-\sin 9^\circ} = \tan 54^\circ \]

Proof

Consider the left-hand side:

\[ \frac{\cos 9^\circ+\sin 9^\circ} {\cos 9^\circ-\sin 9^\circ} \]

Divide numerator and denominator by \[ \cos 9^\circ \]

\[ = \frac{\frac{\cos 9^\circ}{\cos 9^\circ}+\frac{\sin 9^\circ}{\cos 9^\circ}} {\frac{\cos 9^\circ}{\cos 9^\circ}-\frac{\sin 9^\circ}{\cos 9^\circ}} \]

\[ = \frac{1+\tan 9^\circ} {1-\tan 9^\circ} \]

Using the identity:

\[ \tan(45^\circ+\theta) = \frac{1+\tan\theta} {1-\tan\theta} \]

with \[ \theta=9^\circ \]

we get:

\[ = \tan(45^\circ+9^\circ) \]

\[ = \tan 54^\circ \]

Hence,

\[ \frac{\cos 9^\circ+\sin 9^\circ} {\cos 9^\circ-\sin 9^\circ} = \tan 54^\circ \]

Hence proved.