tan ^{-1}left(frac{1+sqrt{3}}{3+sqrt{3}}right | vedclass

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Question

(C) Let $x = 	an ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}ight)$.Simplify the argument: $\frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Let $x = 	an ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}ight)$.Simplify the argument: $\frac{1+\sqrt{3}}{\sqrt{3}(\sqrt{3}+1)} = \frac{1}{\sqrt{3}}$.So, $x = 	an ^{-1}\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6}$.Let $y = \sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}ight)$.Simplify the argument: $\sqrt{\frac{4(2+\sqrt{3})}{3(2+\sqrt{3})}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.So, $y = \sec ^{-1}\left(\frac{2}{\sqrt{3}}ight) = \frac{\pi}{6}$.Therefore, $x + y = \frac{\pi}{6} + \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$.

$\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)$ is equal to $.........$

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