Question

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

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

Answer 1

Answer: (A)

Given,
$asi{n}^{-1}x-bco{s}^{-1}x=c...(i)$
$\Rightarrow a\left(\frac{\pi }{2}-co{s}^{-1}x\right)-bco{s}^{-1}x=c$
$\Rightarrow \left(a+b\right)co{s}^{-1}x=\frac{a\pi }{2}-c$
$\Rightarrow co{s}^{-1}x=\frac{a\pi /2-c}{a+b}$
Again from Eq. $(i)$
$asi{n}^{-1}x-b(\frac{\pi }{2}-si{n}^{-1}x)=c$
$\Rightarrow (si{n}^{-1}x)-(a+b)=c+\frac{b\pi }{2}$
$\Rightarrow si{n}^{-1}x=\frac{c+b\pi /2}{a+b}$
$\therefore asi{n}^{-1}x+bco{s}^{-1}x=\frac{a(c+b\pi /2)}{a+b}$
$+\frac{b(a\pi /2-c)}{a+b}$
$=\frac{c(a-b)+ab\pi }{a+b}$