JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
\(\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)\) is equal to \(.........\)
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\tan ^{-1}\left(\frac{1+\sqrt{3}}{3+\sqrt{3}}\right)+\sec ^{-1}\left(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}}\right)\) \(=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)+\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{3}\)
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