MCQ
$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $
- A$\frac{1}{2}\tan \theta $
- B$\frac{1}{2}\cot \theta $
- ✓$\tan \theta $
- D$\cot \theta $
✓
Answer
Correct option: C.
$\tan \theta $
c
(c) $\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }}$
(c) $\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }}$
$ = \frac{{\sin \theta + 2\sin \theta \cos \theta }}{{2{{\cos }^2}\theta + \cos \theta }} $
$= \frac{{\sin \theta (1 + 2\cos \theta )}}{{\cos \theta (1 + 2\cos \theta )}} $
$= \tan \theta $.
Trick : Put $\theta = 30^\circ $,
since for $\theta = 30^\circ $ no option will give the common value.
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