1.

\(tan^{-1}\sqrt{3}+sec^{-1}⁡2 – cos^{-1}⁡1\) is equal to ________(a) 0(b) \(\frac{2π}{3}\)(c) \(\frac{π}{3}\)(d) \(\frac{π}{4}\)This question was addressed to me in an international level competition.Question is taken from Inverse Trigonometric Functions Basics topic in chapter Inverse Trigonometric Functions of Mathematics – Class 12

Answer»

Right answer is (B) \(\FRAC{2π}{3}\)

The explanation: \(TAN^{-1}\sqrt{3}=\frac{π}{3}, sec^{-1}⁡2=\frac{π}{3}, COS^{-1}⁡1=0\)

∴\(tan^{-1}\sqrt{3}+sec^{-1}⁡2 -cos^{-1}⁡1=\frac{π}{3}+\frac{π}{3}-0\)

=\(\frac{2π}{3}\).


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