$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)+\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)+\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$ is equal to :

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Inverse Trigonometric Functions

sin^{- 1} (sin \frac{2 π}{3}) + cos^{- 1} (cos \frac{7 π}{6}) + tan^{- 1} (tan \frac{3 π}{4})

is equal to :

Answer: (A)

Using the principal value ranges:

sin^{- 1} (sin \frac{2 π}{3})

: Since

\frac{2 π}{3} \in / [- \frac{π}{2}, \frac{π}{2}]

, we use

sin \frac{2 π}{3} = sin (π - \frac{2 π}{3}) = sin \frac{π}{3}

, so the value is

\frac{π}{3}

cos^{- 1} (cos \frac{7 π}{6})

: Since

\frac{7 π}{6} \in / [0, π]

, we use

cos \frac{7 π}{6} = cos (2 π - \frac{7 π}{6}) = cos \frac{5 π}{6}

, so the value is

\frac{5 π}{6}

tan^{- 1} (tan \frac{3 π}{4})

: Since

\frac{3 π}{4} \in / (- \frac{π}{2}, \frac{π}{2})

, we use

tan \frac{3 π}{4} = tan (\frac{3 π}{4} - π) = tan (- \frac{π}{4})

, so the value is

- \frac{π}{4}

.

Sum:

\frac{π}{3} + \frac{5 π}{6} - \frac{π}{4} = \frac{4 π + 10 π - 3 π}{12} = \frac{11 π}{12}

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option