Question #c070f

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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

Explanation:

to prove
$sec (2 x) + tan (2 x) = \frac{1 + tan (x)}{1 - tan (x)}$ $sec(2x)+tan(2x)=(1+tan(x))/(1-tan(x))$

take LHS

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

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

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

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

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

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

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

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

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

$= \frac{1 + tan (x)}{1 - tan (x)}$ $=(1+tan(x))/(1-tan(x)$

as required.

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Question #c070f

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