MCQ
$\frac{\sin \theta}{1+\cos \theta}$ is equal to
- A$\frac{1-\sin \theta}{\cos \theta}$
- B$\frac{1-\cos \theta}{\cos \theta}$
- ✓$\frac{1-\cos \theta}{\sin \theta}$
- D$\frac{1+\cos \theta}{\sin \theta}$
✓
Answer
Correct option: C.
$\frac{1-\cos \theta}{\sin \theta}$
(C) $\frac{1-\cos \theta}{\sin \theta}$
Explanation: We have, $\frac{\sin \theta}{1+\cos \theta}=\frac{\sin \theta(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$=\frac{\sin \theta(1-\cos \theta)}{1-\cos ^2 \theta}=\frac{\sin \theta(1-\cos \theta)}{\sin ^2 \theta}$
$=\frac{1-\cos \theta}{\sin \theta}$
Explanation: We have, $\frac{\sin \theta}{1+\cos \theta}=\frac{\sin \theta(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)}$
$=\frac{\sin \theta(1-\cos \theta)}{1-\cos ^2 \theta}=\frac{\sin \theta(1-\cos \theta)}{\sin ^2 \theta}$
$=\frac{1-\cos \theta}{\sin \theta}$
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