How do you prove $1 + cos 2 x = \frac{2}{1 + {tan}^{2} x}$ $1+cos2x=2/(1+tan^2x)$ using the double angle identity?

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Answer 1 · 2015-08-13T12:08:06

$cos 2 x = 2 {cos}^{2} x - 1$ $cos 2x = 2cos^2x - 1$
${tan}^{2} x + 1 = {sec}^{2} x$ $tan^2 x + 1 = sec^2 x$

$1 + cos 2 x = 2 {cos}^{2} x = \frac{2}{{sec}^{2} x} = \frac{2}{1 + {tan}^{2} x}$ $1 + cos 2x = 2 cos^2 x = 2/sec^2 x = 2/(1 + tan^2 x)$

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