MCQ
$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$ is equal to
- ✓$\operatorname{cosec}\theta-\cos\theta$
- B$\operatorname{cosec} \theta+\cot \theta$
- C$\operatorname{cosec}^2 \theta+\cot ^2 \theta$
- D$\operatorname{cosec}^2 \theta-\cot ^2 \theta$
✓
Answer
Correct option: A.
$\operatorname{cosec}\theta-\cos\theta$
(a) : We have, $\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}-\sqrt{\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}}$
$=\sqrt{\frac{(1-\cos \theta)^2}{1-\cos ^2 \theta}}$
$
\left[\because(a+b)(a-b)=a^2-b^2\right]
$
$=\sqrt{\frac{(1-\cos \theta)^2}{\sin ^2 \theta}}$
$
\left[\because 1-\cos ^2 \theta=\sin ^2 \theta\right]
$
$=\frac{1-\cos \theta}{\sin \theta}=\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}=\operatorname{cosec} \theta-\cot \theta$
$=\sqrt{\frac{(1-\cos \theta)^2}{1-\cos ^2 \theta}}$
$
\left[\because(a+b)(a-b)=a^2-b^2\right]
$
$=\sqrt{\frac{(1-\cos \theta)^2}{\sin ^2 \theta}}$
$
\left[\because 1-\cos ^2 \theta=\sin ^2 \theta\right]
$
$=\frac{1-\cos \theta}{\sin \theta}=\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}=\operatorname{cosec} \theta-\cot \theta$
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