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