Question

If $a{\sin }^{-1}x-b{\cos }^{-1}x=c,$ then $a{\sin }^{-1}x+b{\cos }^{-1}x$ is equal to

• A. $0$
• B. $\frac{\pi ab+c(b-a)}{a+b}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi ab+c(a-b)}{a+b}$

Answer 1

The correct option is D $\frac{\pi ab+c(a-b)}{a+b}$

$\text{We have}b{\sin }^{-1}x+b{\cos }^{-1}x=\frac{b\pi }{2}\text{and}a{\sin }^{-1}x-b{\cos }^{-1}x=c\text{(given)}\Rightarrow (a+b){\sin }^{-1}x=\frac{\left(b\pi \right)}{2}+c\Rightarrow {\sin }^{-1}x=\frac{\frac{b\pi }{2}+c}{a+b}\text{Similarly}{\cos }^{-1}x=\frac{\frac{\left(a\pi \right)}{2}-c}{a+b}\text{so that}a{\sin }^{-1}x+b{\cos }^{-1}x=\frac{\pi ab+c(a-b)}{a+b}$