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प्रश्न
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उत्तर
\[Here, \]
\[\tan\left( 82 . 5 \right)° = \tan\left( 90 - 7 . 5 \right)°\]
\[ = \cot\left( 7 . 5 \right)°\]
\[ = \frac{1}{\tan\left( 7 . 5 \right)°}\]
\[\text{ We know } , \]
\[\tan\left( \frac{x}{2} \right) = \frac{\text{ sin } x}{1 + \text{ cos } x}\]
\[\text{ On putting } x = 15° , \text{ we get } \]
\[\tan \left( \frac{15}{2} \right)^°= \frac{\sin15°}{1 + \cos15°}\]
\[ = \frac{\sin\left( 45 - 30 \right)° }{1 + \cos\left( 45 - 30 \right)°}\]
\[ = \frac{\sin45° \cos30° - \sin30° \cos45° }{1 + \cos45° \cos30° + \sin45° \sin30° }\]
\[ = \frac{\left( \frac{1}{\sqrt{2}} \right) \times \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \times \left( \frac{1}{\sqrt{2}} \right)}{1 + \left( \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \times \frac{1}{2} \right)}\]
\[ = \frac{\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}}{1 + \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}}\]
\[ = \frac{\sqrt{3} - 1}{2\sqrt{2} + \sqrt{3} + 1}\]
\[\text{ Now } , \]
\[\tan\left( 82 . 5 \right)° = \frac{1}{\tan\left( 7 . 5 \right)° }\]
\[ = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}\]
\[ = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\]
\[ = \frac{\sqrt{3} + 1\left( 2\sqrt{2} + \sqrt{3} + 1 \right)}{\left( \sqrt{3} \right)^2 - 1^2}\]
\[ = \frac{2\sqrt{6} + 3 + \sqrt{3} + 2\sqrt{2} + \sqrt{3} + 1}{3 - 1}\]
\[ = \frac{2\sqrt{6} + 2\sqrt{3} + 2\sqrt{2} + 4}{2}\]
\[ = \sqrt{6} + \sqrt{3} + \sqrt{2} + 2\]
\[ = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} . . . \left( 1 \right)\]
\[ = \sqrt{6} + \sqrt{3} + 2 + \sqrt{2}\]
\[ = \sqrt{3}\left( \sqrt{2} + 1 \right) + \sqrt{2}\left( \sqrt{2} + 1 \right)\]
\[ = \left( \sqrt{3} + \sqrt{2} \right)\left( \sqrt{2} + 1 \right) . . . \left( 2 \right)\]
\[\text{ From eqs } . \left( 1 \right) \text{ and } \left( 2 \right), \text{ we get} \]
\[ \tan\left( 82 . 5 \right)° = \left( \sqrt{3} + \sqrt{2} \right)\left( \sqrt{2} + 1 \right) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}\]
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