Question

If $asi{n}^{-1}x-bco{s}^{-1}x=c$, then $asi{n}^{-1}x+bco{s}^{-1}x$ is equal to

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

Answer 1

Answer: (D)

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

We have $bsi{n}^{-1}x+bco{s}^{-1}x=\frac{b\pi }{2}$
and $asi{n}^{-1}x-bco{s}^{-1}x=c$ (given)
$\Rightarrow (a+b)si{n}^{-1}x=\frac{b}{2}\pi +c$
$\Rightarrow si{n}^{-1}x=\frac{\left(b\pi \right)/2+c}{a+b}$
Similarly $co{s}^{-1}x=\frac{\left(a\pi \right)/2-c}{a+b}$
so that $asi{n}^{-1}x+bco{s}^{-1}x=\frac{\pi ab+c(a-b)}{a+b}$