MCQ
$(\operatorname{cosec} \theta-\cot \theta)^2=?$
- A$\frac{1+\sin \theta}{1-\sin \theta}$
- ✓$\frac{1-\cos \theta}{1+\cos \theta}$
- C$\frac{1-\sin \theta}{1+\sin \theta}$
- D$\frac{1+\cos \theta}{1-\cos \theta}$
✓
Answer
Correct option: B.
$\frac{1-\cos \theta}{1+\cos \theta}$
(b) $\frac{1-\cos \theta}{1+\cos \theta}$
Explanation: $(\operatorname{cosec} \theta-\cot \theta)^2=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^2=\frac{(1-\cos \theta)^2}{\sin ^2 \theta}=\frac{(1-\cos \theta)^2}{\left(1-\cos ^2 \theta\right)}=\frac{(1-\cos \theta)}{(1+\cos \theta)}$
Explanation: $(\operatorname{cosec} \theta-\cot \theta)^2=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^2=\frac{(1-\cos \theta)^2}{\sin ^2 \theta}=\frac{(1-\cos \theta)^2}{\left(1-\cos ^2 \theta\right)}=\frac{(1-\cos \theta)}{(1+\cos \theta)}$
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