1) First, let us write tangent in terms of exponentials using "\\sin x = \\frac{e^{ix} - e^{-i x}}{2 i}", "\\cos x = \\frac{e^{i x} + e^{-i x}}{2}":
"\\tan x = \\frac{\\sin x}{\\cos x} = \\frac{e^{i x} - e^{-i x}}{i(e^{i x} + e^{-i x})}" .
Using last equation, obtain:"1 + \\tan^2 x = 1 - \\frac{(e^{i x} - e^{-i x})^2}{(e^{i x} + e^{-i x})^2} = \\frac{(e^{i x} - e^{-i x})^2-(e^{i x}+e^{-i x})^2}{(e^{i x} + e^{-i x})^2} = \\frac{2 e^{i x} 2e^{-ix}}{4 \\cos^2 x} = \\frac{1}{\\cos^2x} = \\sec^2 x"
(Here we used "i^2 = -1", "(a-b)(a+b) = a^2 - b^2" .
2) "1 = e^{ix} e^{-i x} = (\\cos x + i \\sin x)(\\cos x - i \\sin x) = \\sin^2 x + \\cos^2 x"
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