If x = r cos A sin B, y = r cos A cosB and z = r sinA, then prove that

1.	If x = r cos A sin B, y = r cos A cosB and z = r sinA, then prove that x^2+y^2+z^2=r^2
Answer» Note: sin²x + cos²x = 1 Step-by-step EXPLANATION: SQUARE on both sides of X, y and z: x² = (rcosAsinB)² = r²cos²Asin²B y² = (rcosAcosB)² = r²cos²Acos²B z² = (rsinA)² = r²sin²A ADD x² and y²: => x² + y² = r²cos²Asin²B + r²cos²Acos²B = r²cos²A(sin²B + cos²B) = r²cos²A(1) = r²cos²A Now, adding (x² + y²) and z²: => x² + y² + z² = r²cos²A + r²sin²A => x² + y² + z² = r²(cos²A + sin²A) = r²(1) => x² + y² + z² = r², as desired

If x = r cos A sin B, y = r cos A cosB and z = r sinA, then prove that x^2+y^2+z^2=r^2

Answer»

Note: sin²x + cos²x = 1

Step-by-step EXPLANATION:

SQUARE on both sides of X, y and z:

x² = (rcosAsinB)² = r²cos²Asin²B

y² = (rcosAcosB)² = r²cos²Acos²B

z² = (rsinA)² = r²sin²A

ADD x² and y²:

=> x² + y² = r²cos²Asin²B + r²cos²Acos²B

= r²cos²A(sin²B + cos²B)

= r²cos²A(1)

= r²cos²A

Now, adding (x² + y²) and z²:

=> x² + y² + z² = r²cos²A + r²sin²A

=> x² + y² + z² = r²(cos²A + sin²A) = r²(1)

=> x² + y² + z² = r², as desired