Question

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

Answer 1

what should you prove ?
if x^2+y^2+z^2=R^2
x^2+y^2+z^2=( R.sinA.cosB)^2 + (R.sinA.sinB)^2+(R.cosA)^2
R^2 is common. After expanding ,
=R^2[(sinA^2)(cosB^2 + sinB^2) + (cosA^2)]
Use the Identity cos^2 + sin^2 = 1..
=R^2 [(sinA^2)*1 + cosA^2)
=R^2( sinA^2 + cosA^2)
=R^2 * 1
=R^2..
hence proved.

Answer 2

$\bold {x^2+y^2+z^2=r^2}$

Step-by-step explanation:

Given,

x = r sin A cos B, y = r sin A sin B and z = r cos A

we have to prove,

$x^2+y^2+z^2=r^2$

Now,

$x^2+y^2+z^2= ( r.sinA.cosB)^2 + (r.sinA.sinB)^2+(r.cosA)^2$

                           = $r^2[(sin^2A)(cos^2B + sin^2B) + (cos^2A)]$

By using the Identity $cos^2A + sin^2A = 1$

                          = $r^2 [(sin^2A)*1 + cos^2A)$

                         = $r^2( sin^2A + cos^2A)$

                         = $r^2 * 1$

                       $=r^2$

⇒$\bold {x^2+y^2+z^2=r^2}$

Hence proved.

*refer the givrn link for complete question:

https://brainly.in/question/4602202