Question:
Prove that $\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\tan 54^{\circ}$
Solution:
First we will take out cos9°common from both numerator and denominator,
$\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\frac{\cos 9\left(1+\tan 9^{\circ}\right)}{\cos 9^{\circ}\left(1-\tan 9^{\circ}\right)} \Rightarrow \frac{\tan 45^{\circ}+\tan 9^{\circ}}{1-\tan 45^{\circ} \cdot \tan 9^{\circ}}=\tan \left(45^{\circ}+9^{\circ}\right) \Rightarrow \tan 54$
$\left(\right.$ usingtan $\left.(\mathrm{x}+\mathrm{y})=\frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{x} \cdot \tan \mathrm{y}} \operatorname{andtan} 45^{\circ}=1\right)$