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प्रश्न
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
उत्तर
\[\text{ We have } , \]
\[\sqrt{\frac{1 - \cos2x}{1 + \cos2x}} = \sqrt{\frac{2 \sin^2 x}{2 \cos^2 x}}\]
\[ = \frac{\left| \text{ sin } x \right|}{\left| \text{ cos } x \right|}\]
\[ = \frac{\left| \text{ sin } x \right|}{\left| \text{ cos } x \right|} \]
\[ = \frac{- \text{ sin } x}{- \text{ cos } x} \left( \because \pi < x < \frac{3\pi}{2} \right) \]
\[ \therefore \sqrt{\frac{1 - \cos2x}{1 + \cos2x}} = \text{ tan } x\]
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