Express 1/(cos θ − i sin θ) in the form of a + ib and hence prove that
cosθ+isinθ/ cos θ − i sinθ = cos2θ+isin2θ.
"=\\dfrac{\\cos \\theta + i \\sin \\theta}{\\cos^2 \\theta + \\sin^2 \\theta}=\\dfrac{\\cos \\theta + i \\sin \\theta}{1}"
"=\\cos \\theta + i \\sin \\theta"
"\\dfrac{\\cos \\theta + i \\sin \\theta}{\\cos \\theta - i \\sin \\theta}=(\\cos \\theta + i \\sin \\theta)(\\cos \\theta + i \\sin \\theta)"
"=\\cos^2 \\theta+i\\cos \\theta\\sin \\theta+i\\sin \\theta\\cos \\theta-\\sin^2 \\theta"
"=(\\cos^2 \\theta-\\sin^2 \\theta)+i(2\\cos \\theta\\sin \\theta)"
"=\\cos(2\\theta)+i\\sin(2\\theta)"
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