If sin⁻¹x + sin⁻¹y + sin⁻¹z = 3π/2, then x⁹⁹ + y⁹⁹ + z⁹⁹ − 1/(x⁹⁹ y⁹⁹ z⁹⁹) equals:
If sin⁻¹x + sin⁻¹y + sin⁻¹z = 3π/2, then x⁹⁹ + y⁹⁹ + z⁹⁹ − 1/(x⁹⁹ y⁹⁹ z⁹⁹) equals:
- A. 0
- B. 1
- C. 2
- D. −1
Answer: C) 2
Explanation: Maximum value of sin⁻¹ is π/2. For sum to be 3π/2, each must be π/2. So x=y=z=1. Then 1+1+1 − 1/1 = 3−1=2.
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