Question

If $asi{n}^{-1}x-bco{s}^{-1}x=c,thenasi{n}^{-1}x+bco{s}^{-1}x$ 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

Answer: (D)

$\frac{\pi ab+c(a-b)}{a+b}$

$bco{s}^{-1}\left(x\right)+bsi{n}^{-1}\left(x\right)=\frac{b\pi }{2}$

$[\therefore co{s}^{-1}(x)+si{n}^{-1}(x)=\frac{\pi }{2}]$

$asi{n}^{-1}\left(x\right)-bco{s}^{-1}\left(x\right)=c$

Adding both the equations, we get

$si{n}^{-1}\left(x\right)=\frac{b\pi +2c}{2(a+b)}$

Similarly

$co{s}^{-1}\left(x\right)=\frac{a\pi -2c}{2(a+b)}$

Hence

$asi{n}^{-1}\left(x\right)+bco{s}^{-1}\left(x\right)$

$=\frac{2\pi ab+2c(a-b)}{2(a+b)}$

$=\frac{\pi \left(ab\right)+c(a-b)}{a+b}$