Math, asked by JatinAastha, 1 year ago

If x= r sinAcosC, y= r sinAsinC and z= r cosA. Prove that r^2 = x^2+y^2+z^2.

Answers

Answered by sagarvvrv
5
(x^2)+(y^2)+(z^2)=((r^2)(sinA)^2(cosC)^2)+((r^2)(sinA)^2(sinC)^2)+((r^2)(cosA)^2)

Take r^2 common on rhs

then rhs= (r^2)((sinA)^2(cosC)^2+(sinA)^2(sinC)^2+(cosA)^2)

taking (sinA)^2 common
rhs= (r^2)[((sinA)^2)((sinC)^2+(cosC)^2)+(cosA)^2]
=(r^2)[((sinA)^2)+((cosA)^2)]
=r^2
hence,proved
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