tan - 1 ( 1 + 3 3 + 3 ) + sec - 1 ( 8 + 4 3 6 + 3 3 ) is equal to [2023]

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Solution

Q.
$\tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$ is equal to [2023]

1 $\frac{π}{2}$

2 $\frac{π}{4}$

3 $\frac{π}{6}$

4 $\frac{π}{3}$

Ans.
(4)

$\tan^{- 1} (\frac{1 + \sqrt{3}}{3 + \sqrt{3}}) + \sec^{- 1} (\sqrt{\frac{8 + 4 \sqrt{3}}{6 + 3 \sqrt{3}}})$

$\tan^{- 1} (\frac{1 + \sqrt{3}}{\sqrt{3} (\sqrt{3} + 1)}) + \tan^{- 1} (\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{6 + 3 \sqrt{3}}})$

$\tan^{- 1} (\frac{1}{\sqrt{3}}) + \tan^{- 1} (\frac{\sqrt{2 + \sqrt{3}}}{\sqrt{3 (2 + \sqrt{3})}})$

$\tan^{- 1} (\frac{1}{\sqrt{3}}) + \tan^{- 1} (\frac{1}{\sqrt{3}}) = \frac{π}{6} + \frac{π}{6} = \frac{π}{3}$

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