Question

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

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (A)

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