Prove that
Question:

Prove that $\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{0}+\sin 8^{0}}=\tan 37^{0}$

Solution:

First we will take out cos8° common from both numerator and denominator,

$\frac{\cos 8^{\circ}-\sin 8^{\circ}}{\cos 8^{\circ}+\sin 8^{\circ}}=\frac{\cos 8^{\circ}\left(1-\tan 8^{\circ}\right)}{\cos 8^{\circ}\left(1+\tan 8^{\circ}\right)} \Rightarrow \frac{\tan 45^{\circ}-\tan 8^{\circ}}{1+\tan 45^{\circ} \cdot \tan 8^{\circ}}=\tan \left(45^{\circ}-8^{\circ}\right) \Rightarrow \tan 37^{\circ}$

$\left[\right.$ using $\tan (\mathrm{x}-\mathrm{y})=\frac{\tan \mathrm{x}-\tan \mathrm{y}}{1+\tan \mathrm{x} \cdot \tan \mathrm{y}}$ and $\tan 45^{\circ}=1$