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Prove that (.^(n)C(1)sin2x+.^(n)C(2)sin4...

Prove that `(.^(n)C_(1)sin2x+.^(n)C_(2)sin4x+.^(n)C_(3)sin6x+"…..")/(1+.^(n)C_(1)cos2x+.^(n)C_(2)cos4x+.^(n)C_(3)cos 6x+"……")`

Text Solution

Verified by Experts

L.H.S. `=(underset(r=0)overset(n)sum.^(n)C_(r)sin2rx)/(underset(r=0)overset(n)sum.^(n)C_(r)cos2x)" "(1)`
`0=(underset(r=0)overset(n)sum.^(n)C_(n-r)sin2(n-r)x)/(underset(r=0)overset(n)sum.^(n)C_(n-r)cos2(n-r)x)`
`=(underset(r=0)overset(n)sum.^(n)C_(r)sin(2n-2r)x)/(underset(r=0)overset(n)sum.^(n)C_(r)cos(2n-2r)x)" "(2)`
`=(underset(r=0)overset(n)sum.^(n )C_(r)sin2rx+underset(r=0)overset(n)sum.^(n)C_(r)sin(2n-2r)x)/(underset(r=0)overset(n)sum.^(n)C_(r)cos2rx+underset(r=0)overset(n)sum.^(n)C_(r)cos(2n-2r)x)`.
(Using `a/b = c/d = (a+c)/(b+d)` and using (1) and (2))
`= (underset(r=0)overset(n)sum.^(n)C_(r)[sin2rx+sin(2n-2r)x])/(underset(r=0)overset(n)sum.^(n)C_(r)[cos2rx+cos(2n-2r)x])`
`= (2sin nx underset(r=0)overset(n)sum.^(n)C_(r)cos(2r-n)x)/(2cos nx underset(r=0)overset(n)sum.^(n)C_(r)cos(2r-n)x) = tan nx`
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