Prove that: (cos 8° − sin 8°)/(cos 8° + sin 8°) = tan 37°

Question

Prove that:

\[ \frac{\cos 8^\circ-\sin 8^\circ} {\cos 8^\circ+\sin 8^\circ} = \tan 37^\circ \]

Proof

Consider the left-hand side:

\[ \frac{\cos 8^\circ-\sin 8^\circ} {\cos 8^\circ+\sin 8^\circ} \]

Divide numerator and denominator by \[ \cos 8^\circ \]

\[ = \frac{\frac{\cos 8^\circ}{\cos 8^\circ}-\frac{\sin 8^\circ}{\cos 8^\circ}} {\frac{\cos 8^\circ}{\cos 8^\circ}+\frac{\sin 8^\circ}{\cos 8^\circ}} \]

\[ = \frac{1-\tan 8^\circ} {1+\tan 8^\circ} \]

Using the identity:

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

with \[ \theta=8^\circ \]

we get:

\[ = \tan(45^\circ-8^\circ) \]

\[ = \tan 37^\circ \]

Hence,

\[ \frac{\cos 8^\circ-\sin 8^\circ} {\cos 8^\circ+\sin 8^\circ} = \tan 37^\circ \]

Hence proved.