JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q19

MathematicsSingle correct+4 / -1Question 19 of 90

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

\dfrac{11\pi}{12}

\dfrac{17\pi}{12}

\dfrac{31\pi}{12}

-\dfrac{3\pi}4

Answer and worked solution

Official answer: A

Principal values. $\sin^{-1}\sin\frac{2\pi}3=\frac\pi3$ , $\cos^{-1}\cos\frac{7\pi}6=\frac{5\pi}6$ , $\tan^{-1}\tan\frac{3\pi}4=-\frac\pi4$ .

Sum. $\frac{4\pi+10\pi-3\pi}{12}=\dfrac{11\pi}{12}$ .

(A)

Topics:Inverse trigonometric functions

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 19