If x=rsinAcosC, y=rsinAsinC and z=rcosA, prove . . .

Question : If x=rsinAcosC, y=rsinAsinC and z=rcosA, prove that r²=x²+y²+z².

Doubt by Jaskirat

Solution : 

x=rsinAcosC 

y=rsinAsinC

z=rcosA

r²=x²+y²+z²

RHS

x²+y²+z²

=(rsinAcosC)²+(rsinAsinC)²+(rcosA)²
=r²sin²Acos²C+r²sin²Asin²C+r²cos²A

=r²[sin²Acos²C+sin²Asin²C+cos²A]
=r²[sin²A(cos²C+sin²C)+cos²A]
=r²[sin²A(1)+cos²A] (∵sin²θ+cos²θ=1)

=r²[sin²A+cos²A]
=r²[1] (∵sin²θ+cos²θ=1)

=r²

=LHS

LHS=RHS

Hence Proved.