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# please give me ans for the attached question                       see attach

Aamina
64 Points
4 years ago
m = a cos3 $\theta$ + 3a cos$\theta$ sin2$\theta$

n =  a sin $\theta$ + 3a cos2$\theta$ sin$\theta$

m+n = a [ cos3 $\theta$ + sin3 $\theta$ + 3cos$\theta$ sin2$\theta$ +3 cos2$\theta$ sin$\theta$ ]

m+n =a [ cos$\theta$ +sin$\theta$ ]3

(m+n)2/3 = a2/3 { [cos$\theta$ +sin$\theta$]3 }2/3

(m+n)2/3 = a2/3 [cos$\theta$ +sin$\theta$]2 …..................(1)

(m+n)2/3 = a2/3 [1 + 2sin$\theta$ cos$\theta$]

In the same way,

(m-n)2/3 = a2/3  [1 – 2sin$\theta$cos$\theta$]....................(2)

(m+n)2/3 + (m-n)2/3  = a2/3 [1+2sin$\theta$cos$\theta$ +1- 2sin$\theta$cos$\theta$]

= a2/3 [2]

=2a2/3

=2$\sqrt[]{}$$\dpi{80} \fn_cm 3\sqrt{}$a2

Hence option (D) is the answer.

HOPE THIS HELPS.
VENKATA RAMANA
37 Points
4 years ago
thank u very much frnd...