MCQ
The value of $\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}$ is :
- A$\cot\theta-\text{cosec }\theta$
- ✓$\text{cosec }\theta+\cot\theta$
- C$\text{cosec}^2\theta+\cot^2\theta$
- D$(\cot\theta+\text{cosec }\theta)^2$
✓
Answer
Correct option: B.
$\text{cosec }\theta+\cot\theta$
The given expression is $\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}$
Multiplying both the numerator and denominator under the root by $(1+\cos\theta)$, we have
$\sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{(1-\cos^2\theta)}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}}$
$=\frac{1+\cos\theta}{\sin\theta}$
$=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}$
$=\text{cosec }\theta+\cot\theta$
Therefore, the correcr choise is $(b).$
Multiplying both the numerator and denominator under the root by $(1+\cos\theta)$, we have
$\sqrt{\frac{(1+\cos\theta)(1+\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{(1-\cos^2\theta)}}$
$=\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}}$
$=\frac{1+\cos\theta}{\sin\theta}$
$=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}$
$=\text{cosec }\theta+\cot\theta$
Therefore, the correcr choise is $(b).$
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