MCQ
$1+\frac{\cot ^2 \alpha}{1+\operatorname{cosec} \alpha}=$
- A$\sin \alpha$
- B$\sec \alpha$
- ✓$\operatorname{cosec} \alpha$
- D$\tan \alpha$
✓
Answer
Correct option: C.
$\operatorname{cosec} \alpha$
$=1+\frac{\cot ^2 \alpha}{1+\operatorname{cosec} \alpha}$
$=1+\frac{\operatorname{cosec}^2 \alpha-1}{1+\operatorname{cosec\alpha }}$
$=1+\frac{(\operatorname{cosec} \alpha-1)(\operatorname{cosec} \alpha+1)}{1+\operatorname{cosec} \alpha}$
$=1+\operatorname{cosec} \alpha-1$
$=\operatorname{cosec} \alpha$
$=1+\frac{\operatorname{cosec}^2 \alpha-1}{1+\operatorname{cosec\alpha }}$
$=1+\frac{(\operatorname{cosec} \alpha-1)(\operatorname{cosec} \alpha+1)}{1+\operatorname{cosec} \alpha}$
$=1+\operatorname{cosec} \alpha-1$
$=\operatorname{cosec} \alpha$
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