Question

If $asi{n}^{-1}x-bco{s}^{-1}x=c$, then the value of $asi{n}^{-1}x+bco{s}^{-1}x$ (whenever 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

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

As we know that

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

Given $a{\sin }^{-1}x-b{\cos }^{-1}x=c$

$⟹a{\sin }^{-1}x-b(\frac{\pi }{2}-{\sin }^{-1}x)=c$

$⟹(a+b){\sin }^{-1}x=c+\frac{b\pi }{2}$

$⟹{\sin }^{-1}x=\frac{2c+b\pi }{2(a+b)}$

$a{\sin }^{-1}x+b{\cos }^{-1}x=a{\sin }^{-1}x+b(\frac{\pi }{2}-{\sin }^{-1}x)$

$=(a-b){\sin }^{-1}x+b\frac{\pi }{2}$

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

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

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