Express 1/(cosθ − i sinθ) in the form of a + ib and hence prove that
cos θ + i sinθ
cos θ − i sinθ = cos 2θ + i sin2θ.
"\\frac{1}{cos\\theta -isin\\theta}=\\frac{cos\\theta +isin\\theta}{(cos\\theta -isin\\theta)(cos\\theta +isin\\theta)}=\\frac{cos\\theta +isin\\theta}{cos^2\\theta +sin^2\\theta}=cos\\theta +isin\\theta"
"\\frac{cos\\theta +isin\\theta}{cos\\theta -isin\\theta}=(cos\\theta +isin\\theta)^2=cos^2\\theta +2isin\\theta cos\\theta-sin^2\\theta= cos 2\u03b8 + i sin2\u03b8"
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