Given y= sin kx/ 1+cos kx, where k is a positive integer, show sinkx d2y/dx2=k^2 y^2?

y = sinkx / (1 + coskx)

dy/dx = (d/dx)[sinkx / (1 + coskx)]
= {(1 + coskx) * [d(sinkx)/dx] - sinkx * [d(1 + coskx)/dx]} / (1 + coskx)²
= {(1 + coskx) * (k coskx) - sinkx * (-sinkx)} / (1 + coskx)²
= {k coskx + k cos²kx + k sin²kx} / (1 + coskx)²
= {k coskx + k (cos²kx + sin²kx)} / (1 + coskx)²
= {k coskx + k} / (1 + coskx)²
= k (coskx + 1) / (1 + coskx)²
= k (1 + coskx) / (1 + coskx)²
= k / (1 + coskx)

d²y/dx² = (d/dx) [k / (1 + coskx)]
= k * [d(1 + coskx)⁻¹/dx]
= k * [d(1 + coskx)⁻¹/(1 + coskx)] * [d(1 + coskx)/d(kx)] * [d(kx)/dx]
= k * [-(1 + coskx)⁻²] * (-sinkx) * k
= k² sinkx / (1 + coskx)²

L.H.S. = sinkx (d²y/dx²)
= sinkx * [k² sinkx / (1 + coskx)²]
= k² sin²kx / (1 + coskx)²
= k² [sinkx / (1 + coskx)]²
= k² y²
= R.H.S.

Hence, sinkx (d²y/dx²) = k² y²