Question

If $sin^{-1} x + sin^{-1} y + sin^{-1}z = \frac{\pi}{2}$ then value of $x^2 + y^2 + z^2 + 2xyz$ equals

• A. $2$
• B. $0$
• C. $-1$
• D. $1$

Answer 1

Answer: (D)

$sin^{-1} x + sin^{-1} z = \frac{\pi}{2} - sin^{-1}y$
Using $cos$ on both sides, we have
$\sqrt{1 - x^2} \sqrt{1 - z^2} = xz + y$
$\Rightarrow x^2 + y^2 + z^2 + 2xyz = 1$
(squaring and adjusting the terms)
Short Cut Method :
$sin^{-1} x + sin^{-1}y + sin^{-1} z = \frac{\pi}{2}$
$\Rightarrow sin^{-1} x = sin^{-1}y = sin^{-1} z = \frac{\pi}{2} \times \frac{1}{3}$
$x = y = z = \frac{1}{2}$
$\therefore x^2 + y^2 + z^2 + 2xyz$
$= \frac{3}{4} + \frac{1}{4} = 1$