If 2 sin^{-1} x - 3 cos^{-1} x = 4, then 2 si | vedclass

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(A) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.Let $A = \sin^{-1} x$ and $B = \cos^{-1} x$. Then $A + B = \frac{\pi}{2}$, which implies

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$\frac{6\pi - 4}{5}$

Vedclass · Answer

(A) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$.Let $A = \sin^{-1} x$ and $B = \cos^{-1} x$. Then $A + B = \frac{\pi}{2}$, which implies $B = \frac{\pi}{2} - A$.The given equation is $2A - 3B = 4$.Substitute $B = \frac{\pi}{2} - A$ into the equation: $2A - 3(\frac{\pi}{2} - A) = 4$.$2A - \frac{3\pi}{2} + 3A = 4 \implies 5A = 4 + \frac{3\pi}{2} = \frac{8 + 3\pi}{2} \implies A = \frac{8 + 3\pi}{10}$.Now find $B$: $B = \frac{\pi}{2} - \frac{8 + 3\pi}{10} = \frac{5\pi - 8 - 3\pi}{10} = \frac{2\pi - 8}{10} = \frac{\pi - 4}{5}$.We need to find $2A + 3B$: $2(\frac{8 + 3\pi}{10}) + 3(\frac{2\pi - 8}{10}) = \frac{16 + 6\pi + 6\pi - 24}{10} = \frac{12\pi - 8}{10} = \frac{6\pi - 4}{5}$.

If $2 \sin^{-1} x - 3 \cos^{-1} x = 4$, then $2 \sin^{-1} x + 3 \cos^{-1} x =$

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