Question

If $2si{n}^{-1}x-3co{s}^{-1}x=4$, then $2si{n}^{-1}x+3co{s}^{-1}x$ 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)

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