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  4. The value of tan-1 (√2+√3 -√2-√3/√2+√3 +√2-√3)

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Q. The value of $\tan^{-1} \frac{\sqrt{2+\sqrt{3}} -\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3} } +\sqrt{2-\sqrt{3}}} $

COMEDKCOMEDK 2008Inverse Trigonometric Functions

Solution:

$\tan^{-1} \frac{\sqrt{2+\sqrt{3}} -\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3} } +\sqrt{2-\sqrt{3}}}$
Rationalising, $ = \tan ^{-1} \left(\frac{\sqrt{2+\sqrt{3}} -\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3} } +\sqrt{2-\sqrt{3}}} \times\frac{\sqrt{2+\sqrt{3}} -\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3} } -\sqrt{2-\sqrt{3}}}\right) $
$ = \tan ^{-1} \left( \frac{2+\sqrt{3} +2 - \sqrt{3} -2 \sqrt{2+\sqrt{3}}\sqrt{2-\sqrt{3}}}{2+\sqrt{3} -2+\sqrt{3}}\right) $
$= \tan ^{-1} \left(\frac{4 -2\sqrt{4-3}}{2\sqrt{3}}\right) =\tan ^{-1} \left(\frac{2}{2\sqrt{3}}\right) $
$= \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} $