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Question: \[\frac{\cos 17{^\circ} + \sin 17{^\circ}}{\cos 17{^\circ} - \sin 17{^\circ}} =\]...

cos⁡17∘+sin⁡17∘cos⁡17∘−sin⁡17∘=\frac{\cos 17{^\circ} + \sin 17{^\circ}}{\cos 17{^\circ} - \sin 17{^\circ}} =

A

tan⁡62∘\tan 62{^\circ}

B

tan⁡56∘\tan 56{^\circ}

C

tan⁡54∘\tan 54{^\circ}

D

tan⁡73∘\tan 73{^\circ}

Answer

tan⁡62∘\tan 62{^\circ}

Explanation

Solution

Divided by cos⁡17o\cos{}17^{o} in numerator and denominator,

we get, cos⁡17o+sin⁡17ocos⁡17o−sin⁡17o\frac{\cos 17^{o} + \sin 17^{o}}{\cos 17^{o} - \sin 17^{o}}

=1+tan⁡17o1−tan⁡17o=tan⁡45o+tan⁡17o1−tan⁡45otan⁡17o=tan⁡62o= \frac{1 + \tan 17^{o}}{1 - \tan 17^{o}} = \frac{\tan 45^{o} + \tan 17^{o}}{1 - \tan 45^{o}\tan 17^{o}} = \tan{}62^{o}.