X=rsinAcosC, y=rsinAsinC, z=rcosA then prove that x 2 +y 2 +z 2 =r 2
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Given, x=rsinAcosC ..equation..1
y=rsinAsinC .....equation 2
z=rcosA .....equation 3
squaring and adding all three equations we get the following
x2 +y2+z2=r2(sin2Acos2C + sin2Asin2C + cos2A)
=r2 {sin2A(cos2C + sin2C) + cos2A}
=r2 {sin2A+ cos2A}
∴x2 +y2+z2=r2
y=rsinAsinC .....equation 2
z=rcosA .....equation 3
squaring and adding all three equations we get the following
x2 +y2+z2=r2(sin2Acos2C + sin2Asin2C + cos2A)
=r2 {sin2A(cos2C + sin2C) + cos2A}
=r2 {sin2A+ cos2A}
∴x2 +y2+z2=r2
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