How do you simplify $\frac{cos 2 x + {sin}^{2} x}{{cos}^{2} x}$ $(cos2x+sin^2x)/cos^2x$ ?

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Answer 1 · 2018-04-20T02:05:23

$1$ $1$

Recall the identity $cos 2 x = {cos}^{2} x - {sin}^{2} x$ $cos2x=cos^2x-sin^2x$ . Applying it, we get

$(cos^2xcancel(-sin^2x+sin^2x))/cos^2x=cancel(cos^2x)/cancel(cos^2x)=1$

How do you simplify (cos2x+sin^2x)/cos^2xcos2x+sin2xcos2x(cos2x+sin^2x)/cos^2x?