# Tan 82 1 ° 2 = ( √ 3 + √ 2 ) ( √ 2 + 1 ) = √ 2 + √ 3 + √ 4 + √ 6 - Mathematics

Numerical
$\tan 82\frac{1° }{2} = \left( \sqrt{3} + \sqrt{2} \right) \left( \sqrt{2} + 1 \right) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}$

#### Solution

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

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.1 | Q 26 | Page 29