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Q.

 If x=rcosαcosβcosγ ; y= r cosαcosβsinγ ; z=rsinαcosβ, μ=rsinβ then  x2+y2+z2+μ2=

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a

r2

b

4r2

c

r

d

2r

answer is C.

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Detailed Solution

x=r cos α cos β cos γ y=r cos α cos β sin γ z=r sin α cos β μ=r sin β

  x2+y2+z2+μ2 =r2 cos2α cos2β cos2γ+r2 cos2α cos2β sin2γ        +r2 sin2α cos2β+r2 sin2β =r2 cos2α cos2βcos2γ+sin2γ+r2 sin2α cos2β+r2 sin2β =r2 cos2α cos2β+r2+sin2α cos2β+r2 sin2β =r2 cos2βcos2α+sin2α+r2sin2β =r2 cos2β+r2 sin2β =r2 cos2β+sin2β =r2

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