Question

Prove the following:

${\displaystyle \frac{{\cos 27}^{\circ }+{\sin 27}^{\circ }}{{\cos 27}^{\circ }-{\sin 27}^{\circ }}}$ = tan72°

Answer 1

L.H.S. = ${\displaystyle \frac{{\cos 27}^{\circ }+{\sin 27}^{\circ }}{{\cos 27}^{\circ }-{\sin 27}^{\circ }}}$ 

= ${\displaystyle \frac{\frac{{\cos 27}^{\circ }}{{\cos 27}^{\circ }}+\frac{{\sin 27}^{\circ }}{{\cos 27}^{\circ }}}{\frac{{\cos 27}^{\circ }}{{\cos 27}^{\circ }}-\frac{{\sin 27}^{\circ }}{{\cos 27}^{\circ }}}}$

= ${\displaystyle \frac{1+{\tan 27}^{\circ }}{1-{\tan 27}^{\circ }}}$

= ${\displaystyle \frac{{\tan 45}^{\circ }+{\tan 27}^{\circ }}{1-{\tan 45}^{\circ }\times {\tan 27}^{\circ }}}$   ...[∵  tan45° = 1]

= tan(45° + 27°)

= tan72°

= R.H.S.