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