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प्रश्न
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
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उत्तर
We have,
\[\sqrt{2 + \sqrt{2 + 2\cos2x}} = \sqrt{2 + \sqrt{2\left( 1 + \cos2x \right)}} \]
\[ = \sqrt{2 + \sqrt{2 . 2 \cos^2 x}}\]
\[ = \sqrt{2 + 2\left| \text{ cos } x \right|}\]
\[ = \sqrt{2 - 2\text{ cos } x} \left( \because \frac{\pi}{2} < x < \pi \right) \]
\[ = \sqrt{2\left( 1 - \text{ cos } x \right)}\]
\[ = \sqrt{2 . 2 \sin^2 \frac{x}{2}}\]
\[ = 2\left| \sin\frac{x}{2} \right|\]
\[ = 2\sin\frac{x}{2} \left( \because \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2} \right)\]
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