Inverse Trigonometric Functions Ans 1
Q1 - 24 January - Shift 1
$\tan ^{-1}(\frac{1+\sqrt{3}}{3+\sqrt{3}})+\sec ^{-1}(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}})$ is equal to
(1) $\frac{\pi}{4}$
(2) $\frac{\pi}{2}$
(3) $\frac{\pi}{3}$
(4) $\frac{\pi}{6}$
Show Answer
Answer: (3)
Solution:
Formula: Simplification
$\tan ^{-1}(\frac{1+\sqrt{3}}{3+\sqrt{3}})+\sec ^{-1}(\sqrt{\frac{8+4 \sqrt{3}}{6+3 \sqrt{3}}})$
$=\tan ^{-1}(\frac{1}{\sqrt{3}})+\sec ^{-1}(\frac{2}{\sqrt{3}})=\frac{\pi}{3}$
