MCQ
If $\pi<\text{x}<2\pi,$ then $\sqrt{\frac{1+\cos\text{x}}{1-\cos\text{x}}}$ is equal to:
- A$\text{cosec x}+\cot \text{x}$
- B$\text{cosec x}-\cot \text{x}$
- C$-\text{cosec x}+\cot \text{x}$
- D$-\text{cosec x}-\cot \text{x}$
Solution:
$-\text{cosec}\text{x} -\cot\text{x}$
$=\sqrt{\frac{1+\cos\text{x}}{1-\cos\text{x}}}$
$=\sqrt{\frac{(1+\cos\text{x})(1+\cos\text{x})}{(1+\cos\text{x})(1+\cos\text{x})}}$
$=\sqrt{\frac{(1+\cos\text{x})^2}{1-\cos^2\text{x}}}$
$=\sqrt{\frac{(1+\cos\text{x})^2}{\sin^2\text{x}}}$
$=\frac{(1+\cos\text{x})}{-\sin\text{x}}$ $ [\text{as},\pi<\text{x}<2\pi,\text{ so }\sin\text{x} \text{ will}\text{ be}\text{ negative}]$
$=-(\text{cosec}\text{ x}+\cot\text{x})$
$= -\text{cosec}\text{ x } -\cot\text{x}$
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