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