How do you verify $1 + tan x tan 2 x = sec 2 x$ $1 + tanxtan2x = sec2x$ ?

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Answer 1 · 2015-06-06T20:36:49

$\frac{1 + sin x . sin 2 x}{cos x . cos 2 x} =$ $(1 + sin x.sin 2x)/(cos x.cos 2x) =$

$= \frac{cos x . cos 2 x + sin x . sin 2 x}{cos x . cos 2 x} =$ $= (cos x.cos 2x + sin x.sin 2x)/(cos x.cos 2x) =$

$= \frac{cos (2 x - x)}{cos x . cos 2 x} = \frac{1}{cos (2 x)}$ $= cos (2x - x)/(cos x.cos 2x) = 1/cos (2x)$ = sec 2x

How do you verify 1 + tanxtan2x = sec2x1+tanxtan2x=sec2x 1 + tanxtan2x = sec2x?