Question

If x = rsinαcosβ , y = rsinαcosβ and z = rcosα, prove that x² + y² + z² = r²​

Answer 1

Answer:

x = rsinAcosC

y= rsinAsinC

z= rcosA

squaring and adding all,

x² +y²+z² =( rsinAcosC)²+(rsinAsinC)+(rcosA)²

              =r²[ sin²Acos²C + sin²Asin²C + cos²A]

              =r² [sin²A(cos²C + sin²C) + cos²A]    ∵cos²α+sin²α =1

              =r²[sin²A+cos²A]    

              =r²

Answer 2

➝ Given :-

⇾x = rsinα.cosβ

⇾y = rsinα.cosβ

⇾z = rcos α

➝ To prove :-

→ x²+y²+z² = r²

➝ Solution :-

⟹x² + y² + z²

⟹ r² sin²α cos²β +r²sin²α sin²β +r²cos²α

⟹ r² sin²α (cos²β + sin²β)+r²cos²α

⟹ r² sin²α +r² cos²α [: cos²β +sin²β = 1]

⟹ r² (sin²α + cos²α) = r [: sin²α +cos²α= 1].

Hence, (x² + y² + z²) = r²

Hence proved.