MCQ
$\frac{{\sin 70^\circ + \cos 40^\circ }}{{\cos 70^\circ + \sin 40^\circ }} = $
- A$1$
- B$\frac{1}{{\sqrt 3 }}$
- ✓$\sqrt 3 $
- D$\frac{1}{2}$
✓
Answer
Correct option: C.
$\sqrt 3 $
c
(c) $\frac{{\sin \,\,{{70}^o} + \cos \,\,{{40}^o}}}{{\cos \,\,{{70}^o} + \sin \,\,{{40}^o}}}$
(c) $\frac{{\sin \,\,{{70}^o} + \cos \,\,{{40}^o}}}{{\cos \,\,{{70}^o} + \sin \,\,{{40}^o}}}$
$ = \frac{{\sin 70^\circ + \sin 50^\circ }}{{\sin 20^\circ + \sin 40^\circ }} $
$= \frac{{2\sin 60^\circ \cos 10^\circ }}{{2\sin 30^\circ \cos ( - 10^\circ )}}$
$ = \frac{{\sin \,\,{{60}^o}}}{{\sin \,\,{{30}^o}}} = \frac{{\sqrt 3 }}{2}.\frac{2}{1} = \sqrt 3 $.
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