Q. If $\sin^{- 1}{\alpha + \sin^{- 1}{\beta + \sin^{- 1}{\gamma = \frac{3\pi}{2},}}}\$then $\alpha\beta + \alpha\gamma + \beta\gamma$ is equal to
A) 1
B) 0
C) 3
D) $- 3$
Explanation
(c)
Since, $- \frac{\pi}{2} \leq \sin^{- 1}{x \leq \frac{\pi}{2}}$
$\therefore\sin^{- 1}{\alpha = \frac{\pi}{2},\sin^{- 1}{\beta = \frac{\pi}{2}and\sin^{- 1}{\gamma = \frac{\pi}{2}}}}$
$\therefore\alpha = \beta = \gamma = 1$
Thus , $\alpha\beta + \alpha\gamma + \gamma\beta = 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.