Question

If $\sin^{-1} x + \sin^{-1}y + \sin^{-1} z = \frac{3 \pi}{2}$ then the value of $x^{100} + y^{100} + z^{100} - \frac{ 3}{x^{101} + y^{101} + z^{101}}$ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C)

2

Answer 2

We have $\sin^{-1} x + \sin^{-1}y + \sin^{-1}z = \frac{3\pi}{2}$ and $\sin^{-1} x \le \frac{\pi}{2}$ it is possible only when $\sin^{-1} x = \frac{\pi}{2} \Rightarrow x = 1$ $\sin^{-1} y = \frac{\pi}{2} \Rightarrow y =1 ;$ $\sin^{-1} z = \frac{\pi}{2} \Rightarrow z = 1$ $\therefore x^{100} + y^{100} + z^{100} - \frac{ 3}{x^{101} + y^{101} + z^{101}}$ $= 1 + 1 + 1 - \frac{3}{3}$ $= 3 - 1$ $= 2$