Prove that: (cos 11° + sin 11°)/(cos 11° − sin 11°) = tan 56°

Question

Prove that:

\[ \frac{\cos 11^\circ+\sin 11^\circ} {\cos 11^\circ-\sin 11^\circ} = \tan 56^\circ \]

Proof

Consider the left-hand side:

\[ \frac{\cos 11^\circ+\sin 11^\circ} {\cos 11^\circ-\sin 11^\circ} \]

Divide numerator and denominator by \[ \cos 11^\circ \]

\[ = \frac{\frac{\cos 11^\circ}{\cos 11^\circ}+\frac{\sin 11^\circ}{\cos 11^\circ}} {\frac{\cos 11^\circ}{\cos 11^\circ}-\frac{\sin 11^\circ}{\cos 11^\circ}} \]

\[ = \frac{1+\tan 11^\circ} {1-\tan 11^\circ} \]

Using the identity:

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

with \[ \theta=11^\circ \]

we get:

\[ = \tan(45^\circ+11^\circ) \]

\[ = \tan 56^\circ \]

Hence,

\[ \frac{\cos 11^\circ+\sin 11^\circ} {\cos 11^\circ-\sin 11^\circ} = \tan 56^\circ \]

Hence proved.