Question

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

• A. $\frac{6\pi -4}{5}$
• B. $\frac{4 - 6\pi}{5}$
• C. $\frac{3\pi }{2}$
• D. $0$

Answer 1

Answer: (A)

$2sin^{-1}x - 3cos^{-1}x = 4$
$\Rightarrow 2\left(\frac{\pi}{2}-cos^{-1}\,x\right) - 3cos^{-1}x = 4$
$\Rightarrow \pi - 5cos^{-x} = 4$
$\Rightarrow cos^{-1}x =\frac{\pi - 4}{5}$
$\Rightarrow 3cos^{-1}x = \frac{3\pi -12}{5}\quad\cdots\left(i\right)$
Similarly, $2sin^{-1}x - 3\left(\frac{\pi}{2}-sin^{-1}\,x\right) = 4$
$\Rightarrow 5sin^{-1}x = 4 + \frac{3\pi}{2}$
$\Rightarrow sin^{-1}x = \left(\frac{4}{5}+\frac{3\pi }{2}\right)$
$\Rightarrow 2sin^{-1}x = \left(\frac{8}{5}+\frac{3\pi }{5}\right)\quad\cdots\left(ii\right)$
Adding eq. $\left(i\right)$ and $\left(ii\right)$, we get
$2 sin^{-1} x + 3 cos^{-1}x = \frac{8+3\pi}{5}+\frac{3\pi-12}{5}$
$= \frac{6\pi -4}{5}$