Question

Prove that:

(iii). $\frac{\cos 8^{\circ }-\sin 8^{\circ }}{\cos 8^{\circ }+\sin 8^{\circ }}=\tan {37}^{\circ }$

Answer 1

To prove:

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

$L.H.S.$

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

$=\frac{1-\frac{\sin 8^{\circ }}{\cos 8^{\circ }}}{1+\frac{\sin 8^{\circ }}{\cos 8^{\circ }}}$

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

$=\frac{\tan {45}^{\circ }-\tan 8^{\circ }}{1+\tan {45}^{\circ }\cdot \tan 8^{\circ }}$

$=\tan (45-8{)}^{\circ }$

$=\tan {37}^{\circ }$

Hence proved.