If x = r sin <spa A. x² + y² + z² = r²

B. x² + y²− z² = r²

C. x²− y² + z² = r²

D. z² + y²− x₂ = r²

Given: x = r sin θ cos ϕ, y = r sin θ sin φ and z = r cos θ, 

Solution:

x = r sin θ cos ϕ

Squaring both sides, we get

x² = r² sin² θ cos²ϕ ……….(i)

and y = r sin θ sin ϕ

Squaring both sides, we get

⇒ y² = r² sin² θ sin²ϕ ……….(ii)

z = r cos θ

Squaring both sides, we get

⇒ z² = r² cos² θ ……….(iii)

Adding (i), (ii) and (iii), we get

x² + y² + z² = r² sin² θ cos²ϕ + r² sin² θ sin²ϕ + r² cos² θ

= r² (sin² θ cos²ϕ + sin² θ sin²ϕ + cos² θ)

= r² [sin² θ (cos²ϕ + sin²ϕ) + cos² θ]

∵ sin² θ + cos² θ = 1

= r² [sin² θ + cos² θ] 

 Again apply the identity  sin² θ + cos² θ = 1

= r² 

Hence x² + y² + z² = r²