MCQ
$\frac{{\cos 17^\circ + \sin 17^\circ }}{{\cos 17^\circ - \sin 17^\circ }} = $
  • $\tan 62^\circ $
  • B
    $\tan 56^\circ $
  • C
    $\tan 54^\circ $
  • D
    $\tan 73^\circ $

Answer

Correct option: A.
$\tan 62^\circ $
a
(a) Divided by $\cos \,\,{17^o}$ in numerator and denominator, 

we get, $\frac{{\cos \,\,{{17}^o} + \sin \,\,{{17}^o}}}{{\cos \,\,{{17}^o} - \sin \,\,{{17}^o}}}$

$ = \frac{{1 + \tan \,\,{{17}^o}}}{{1 - \tan \,\,{{17}^o}}} $

$= \frac{{\tan \,\,{{45}^o} + \tan \,\,{{17}^o}}}{{1 - \tan \,\,{{45}^o}\tan \,\,{{17}^o}}} = \tan \,\,{62^o}$.

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