sinx + sin2x + sin3x
_________________ = tan2x

cosx + cos2x + cos3x


and

sin^3x + cos^3x = (1-sinxcosx)(sinx + cosx)

sin A + sin B = 2 sin ½(A + B) cos ½(A - B)

Use the above formula to get

sin3x + sinx = 2 sin 2x cos x

Also: cos A + cos B = 2 cos ½(A + B) cos ½(A - B)

Cos 3x + cos x = 2 cos 2x cos x

Hence

LHS = (2 sin 2x cos x + sin 2x)/(2cos 2x cos x + cos 2x)

= sin 2x(2cos x +1)/cos 2x (2cos x + 1)

= sin 2x/cos 2x

= tan 2x


For the second part, use factorizing for x^3 + y^3 = (x + y) (x^2 + y^2 - xy)

Hey Raman Its M Vipul... u Can Use The Formula Sin3x = 3sinx - 4sinx^3