If ^ -1 x + ^ -1 y + ^ -1 z = (3π)/(2) and f ( 1 ) = 2, f ( p + q )...

$∵ \dfrac{- \pi}{2} \le \sin^{-1}x \le \dfrac{\pi}{2}, \dfrac{- \pi}{2} \le \sin^{-1}y \le \dfrac{\pi}{2}$

and $\dfrac{- \pi}{2} \le \sin^{-1}z \le \dfrac{\pi}{2}$

Given that,

$\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \dfrac{3\pi}{2}$

Which is possible only when

$\sin^{-1}x = \sin^{-1}y = \sin^{-1}z = \dfrac{\pi}{2}$

$⇒ x = y = z = 1$

Put $p = q = 1$

Then, $f ( 2 ) = f ( 1 ) f ( 1 ) = 2 \times 2 = 4$

and put $p = 1, q = 2$

then, $f ( 3 ) = f ( 1 ) f ( 2 )$

$= 2 \cdot {2}^{2} = 8$

$∴ x^{f ( 1 )} + y^{f ( 2 )} + z^{f ( 3 )} - \dfrac{x + y + z}{x^{f ( 1 )} + y^{f ( 2 )} + z^{f ( 3 )}}$

$= 1 + 1 + 1 - \dfrac{3}{1 + 1 + 1} = 3 - 1 = 2$