If sin –1 x + sin –1 y + sin –1 z = π , then prove that x 4 + y 4 + z 4 + 4x 2 y 2 z 2 = 2(x 2 y 2 + y 2 z 2 + z 2 x 2 )
Text Solution
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sin –1 x + sin –1 y + sin –1 z = π
⇒ (sin –1 x + sin –1 y) = π –sin –1 z
⇒ cos (sin –1 x + sin –1 y) = cos( π –sin –1 z)
⇒ cos (sin –1 x) cos(sin –1 y) – xy = – cos sin –1 z
⇒ 
– xy = – cos cos –1 
⇒ 
= xy – 
Squaring both sides we have
⇒ (1–x 2 ) (1–y 2 ) = x 2 y 2 + 1 – z 2 – 2xy 
⇒ x 2 + y 2 – z 2 = 2xy 
Again squaring both sides, we get
x 4 + y 4 + z 4 + 2x 2 y 2 – 2y 2 z 2 – 2z 2 x 2 = 4x 2 y 2 – 4x 2 y 2 z 2 ⇒ x
4 + y 4 + z 4 + 4x 2 y 2 z 2 = 2(x 2 y 2 + y 2 z 2 + z 2 x 2 )
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