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Mathematics Question on Inverse Trigonometric Functions

The value of tan⁡−12+3−2−32+3+2−3\tan^{-1} \frac{\sqrt{2+\sqrt{3}} -\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3} } +\sqrt{2-\sqrt{3}}}

A

π6 \frac{\pi}{6}

B

π4 \frac{\pi}{4}

C

π3 \frac{\pi}{3}

D

π2 \frac{\pi}{2}

Answer

π6 \frac{\pi}{6}

Explanation

Solution

tan⁡−12+3−2−32+3+2−3\tan^{-1} \frac{\sqrt{2+\sqrt{3}} -\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3} } +\sqrt{2-\sqrt{3}}}
Rationalising, =tan⁡−1(2+3−2−32+3+2−3×2+3−2−32+3−2−3)= \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(2+3+2−3−22+32−32+3−2+3)= \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(4−24−323)=tan⁡−1(223)= \tan ^{-1} \left(\frac{4 -2\sqrt{4-3}}{2\sqrt{3}}\right) =\tan ^{-1} \left(\frac{2}{2\sqrt{3}}\right)
=tan⁡−1(13)=π6= \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}