MCQ
If $\frac{\pi}{2}<\text{x}<\frac{3\pi}{2},$ then $\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}$ is equal to:
- A$\sec\text{x}-\tan\text{x}$
- B$\sec\text{x}+\tan\text{x}$
- C$\tan\text{x}-\sec\text{x}$
- Dnone of these
Solution:
$\tan\text{x}-\sec\text{x}$
$\sqrt{\frac{1-\sin\text{x}}{1+\sin\text{x}}}$
$=\sqrt{\frac{(1-\sin\text{x})(1-\sin\text{x})}{(1+\sin\text{x})(1-\sin\text{x})}}$
$=\sqrt{\frac{(1-\sin\text{x})^2}{1-\sin^2\text{x}}}$
$=\sqrt{\frac{(1-\sin\text{x})^2}{\cos^2\text{x}}}$
$=\frac{(1-\sin\text{x})}{-\cos\text{x}}$ $[\text{as},\frac{\pi}{2}<\text{x}<\frac{3\pi}{2},\text{so}\cos\theta \text{ will}\text{ be}\text{ negative}]$
$=-(\sec\text{x}-\tan\text{x})$
$=-\sec\text{x} +\tan\text{x}$
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