Question #c070f

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Answer 1 · 2016-10-09T16:01:11

Using the double angle formulas and combing the fractions and simplifying the algebra

to prove
$sec(2x)+tan(2x)=(1+tan(x))/(1-tan(x))$

take LHS

$sec(2x)+tan(2x)= 1/cos(2x) +2tan(x)/(1-tan^2(x))$

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

Divide every term of the first fraction by $cos^2(x)$

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

$=sec^2(x)/(1-tan^2(x))+2tan(x)/(1-tan^2(x))$

$=(1+tan^2(x))/(1-tan^2(x))+2tan(x)/(1-tan^2(x))$

$=(1+2tan(x)+tan^2(x))/(1-tan^2(x))$

$=(1+tan(x))^2/((1-tan(x))(1+tan(x))$

cancelling a $(1+tan(x))$ bracket

$=(1+tan(x))/(1-tan(x)$

as required.