Question

If $a{\sin }^{-1}x-b{\cos }^{-1}x=c$, then the value of $a{\sin }^{-1}x+b{\cos }^{-1}x$ (whener exists) 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

Answer: (D)

We have $b{\sin }^{-1}x+b{\cos }^{-1}x=\frac{b\pi }{2}$ .....(1)
and ${\sin }^{-1}x-b{\cos }^{-1}x=c$....(2) (given)
$\therefore$ On adding (1) and (2), we get $(a+b){\sin }^{-1}x=\frac{b\pi }{2}+c$
$\Rightarrow {\sin }^{-1}x=\frac{\frac{b\pi }{2}+c}{a+b}$. Similarly ${\cos }^{-1}x=\frac{\frac{a\pi }{2}-c}{a+b}$
Hence $\left(a{\sin }^{-1}x+b{\cos }^{-1}x\right)=\frac{\pi ab+c(a-b)}{a+b}$