Question

If sin^–1x + sin^–1y + sin^–1 z = $\frac { 3 \pi } { 2 }$, then the value of

x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ – $\frac { 9 } { x ^ { 101 } + y ^ { 101 } + z ^ { 101 } }$ is equal to

• A. 0
• B. 3
• C. –3
• D. 9

Answer 1

Answer: (A)

0

Answer 2

sin^–1x + sin^–1y + sin^–1z = $\frac { 3 \pi } { 2 }$

sin^–1x = sin^–1y = sin^–1z =$\frac { \pi } { 2 }$Ž x = y = z = sin $\frac { \pi } { 2 }$ = 1

= x¹⁰⁰ + y¹⁰⁰ + z¹⁰⁰ – $\frac { 9 } { x ^ { 101 } + y ^ { 101 } + z ^ { 101 } }$ = 3 –$\frac { 9 } { 3 }$= 0