Question

If sin^–1x + sin^–1 y + sin^–1 z = p, then x⁴ + y⁴ + z⁴ + 4x²y²z² = k (x²y² + y²z² + z²x²), where k is equal to –

• A. 1
• B. 2
• C. 4
• D. None of these

Answer 1

Answer: (B)

2

Answer 2

We have sin^–1 x + sin^–1 y = p – sin^–1 z

or, x + y = z

or, x² (1– y²) = z² + y² (1 – x²) – 2yz $\sqrt { \left( 1 - x ^ { 2 } \right) }$

or, (x² – z² – y²)² = 4y² z² (1 – x²)

or, x⁴ + y⁴ + z⁴ – 2x² z² + 2y² z² – 2x² y² + 4x² y² z² – 4y² z² =0

or, x⁴ + y⁴ + z⁴ + 4x² y² z² = 2(x² y² + y² z² + z² x²)

\ k = 2.