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प्रश्न
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
उत्तर
\[\text{ We have } , \]
\[\sqrt{1 - \sin2x}\]
\[ = \sqrt{\sin^2 x + \cos^2 x - 2\text{ sin } x \text{ cos } x}\]
\[ = \sqrt{\left( \text{ sin } x - \text{ cos } x \right)^2} \]
\[ = \left| \text{ sin } x - \text{ cos } x \right|\]
\[ = \text{ sin } x - \text{ cos } x\]
\[ \left[ \because \text{ sin } x > \text{ cos } x \text{ for } \frac{\pi}{4} < x < \frac{\pi}{2} \right]\]
\[ \therefore \sqrt{1 - \text{ sin } 2x} = \text{ sin } x - \text{ cos } x\]
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