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

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 $\left(a \sin ^{-1} x+b \cos ^{-1} x\right)=\frac{\pi a b+c(a-b)}{a+b}$