Question

Prove that

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

Answer 1

\[\text{ LHS }= \frac{\cos9^\circ + \sin9^\circ}{\cos9^\circ - \sin9^\circ}\]
\[ = \frac{\frac{\cos9^\circ}{\cos9^\circ} + \frac{\sin9^\circ}{\cos9^\circ}}{\frac{\cos9^\circ}{\cos9^\circ} - \frac{\sin9^\circ}{\cos9^\circ}} \left(\text{ Dividing the numerator and denominator by }\cos9 \right)\]
\[ = \frac{1 + \tan9^\circ}{1 - \tan9^\circ}\]
\[ = \frac{1 + \tan9^\circ}{1 + 1 \times \tan9^\circ}\]
\[ = \frac{\tan45^\circ + \tan9^\circ}{1 - \tan45^\circ \times \tan9^\circ} \left(\text{ As }\tan45^\circ = 1 \right)\]
\[ = \tan\left( 45^\circ + 9^\circ \right) \left[\text{ As }\frac{\tan A + \tan B}{1 - \tan A \tan B} = \tan\left( A + B \right) \right]\]
\[ = \tan54^\circ\]
 = RHS
Hence proved .