Question

if x= 3 cosAcosB, y = 3 cosAsinB and z= 3sinA, prove that x2+y2+z2=9

Answer 1

x = 3 cosA cosB


square on both sides,


x² = 9 cos²A cos²B -----: ( 1 )


y = 3 cosA sinB

square on both sides,

y² = 9 cos²A sin²B -----: ( 2 )


z = 3 sinA

squares on both sides,

z² = 9 sin²A ------: ( 3 )




adding 1 , 2 and 3


x² + y² + z² = 9 cos²A cos²B + 9 cos²A sin²B + 9 sin²A


x² + y² + z² = 9[ cos²A cos²B + cos²A sin²B + sin²A ]


x² + y² + z² = 9 [ cos²A{ cos²B + sin²B } + sin²A ]


x² + y² + z² = 9[ cos²A ( 1 ) + sin²A ]


x² + y² + z² = 9[ cos²A + sin²A ]


x² + y² + z² = 9( 1 )


x² + y² + z² = 9




Proved.
$\:$

Answer 2

Given,

x = 3cosAcosB

y = 3cosAsinB

z = 3sinA

Now,

x =  3cosAcosB

Making square  of  both sides,

x² = 9cos²Acos²B  ........................i)

Similarly,

y² = 9cos²Asin²B...................ii)

z²= 9sin²A ................iii)

Adding  equation i), ii) and iii)

x² + y² + z² = 9cos²Acos²B +  9cos²Asin²B + 9sin²A

x² + y²  + z² = 9cos²A(cos²A + sin²A)  + 9sin²A

x² + y² + z² = 9cos²A×1 + 9sin²A    (∵sin²Ф+cos²Ф=1)

x² + y² + z² = 9cos²A + 9sin²A

x² + y²  + z² = 9(cos²A + sin²A)

x² + y²  + z² = 9 × 1

x² + y²  + z² = 9