MCQ
$(\text{cosec }\theta-\cot\theta)^2=?$
- A$\frac{1+\cos\theta}{1-\cos\theta}$
- ✓$\frac{1-\cos\theta}{1+\cos\theta}$
- C$\frac{1+\sin\theta}{1-\sin\theta}$
- DNone of these.
✓
Answer
Correct option: B.
$\frac{1-\cos\theta}{1+\cos\theta}$
$(\text{cosec }\theta-\cot\theta)^2=\Big(\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}\Big) ^2$
$=\Big(\frac{1-\cos\theta}{\sin\theta}\Big)^2$
$=\frac{(1-\cos\theta)^2}{\sin^2\theta}$
$=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}$
$=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}$
$=\frac{1-\cos\theta}{1+\cos\theta}$
$=\Big(\frac{1-\cos\theta}{\sin\theta}\Big)^2$
$=\frac{(1-\cos\theta)^2}{\sin^2\theta}$
$=\frac{(1-\cos\theta)^2}{1-\cos^2\theta}$
$=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}$
$=\frac{1-\cos\theta}{1+\cos\theta}$
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