MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Use the standard identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$ and treat the inverse trigonometric expressions as variables to solve the linear equations.Explanation:Let $A = \sin^{-1}x$ and $B = \cos^{-1}x$.For the same value of $x$, the identity gives $A + B = \frac{\pi}{2}$.Hence, $A = \frac{\pi}{2} - B$.The given equation is $2A - 3B = 4$.Substitute $A$: $2\left(\frac{\pi}{2} - B\right) - 3B = 4$.Simplify: $\pi - 2B - 3B = 4$, so $\pi - 5B = 4$.Therefore, $B = \frac{\pi - 4}{5}$.We need $2A + 3B$.Using $A = \frac{\pi}{2} - B$: $2A + 3B = 2\left(\frac{\pi}{2} - B\right) + 3B$.This simplifies to $\pi - 2B + 3B = \pi + B$.Substitute $B = \frac{\pi - 4}{5}$: $\pi + \frac{\pi - 4}{5} = \frac{5\pi + \pi - 4}{5}$.Thus, the required value is $\frac{6\pi - 4}{5}$.Answer:$\frac{6\pi - 4}{5}$Option A.