Question

If $ asin^{-1}x - bcos^{-1} x = c $ , $ a sin^{-1}x + b cos^{-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,
$a\, sin^{-1} \,x - b \,cos^{-1} \,x = c\,\,\,...(i)$
$\Rightarrow a\left(\frac{\pi}{2} - cos^{-1}x\right) - b\, cos^{-1} x=c $
$\Rightarrow \left(a + b\right)cos^{-1}x = \frac{a\pi}{2} - c $
$\Rightarrow cos^{-1} x = \frac{a\pi /2 -c}{a+b}$
Again from Eq. $(i)$
$a\, sin^{-1} x - b (\frac{\pi}{2} - sin^{-1} \,x ) = c$
$\Rightarrow (sin^{-1} x) - ( a + b ) = c + \frac{b\pi}{2}$
$\Rightarrow sin^{-1} x = \frac{c + b\pi/2}{a + b}$
$\therefore a\, sin^{-1} \,x + b \,cos^{-1}\,x = \frac{a(c + b\pi/2)}{a + b}$
$ + \frac{b(a\pi/2 - c )}{a + b}$
$ = \frac{c(a - b) + ab\pi}{a + b}$