# Question c070f

Oct 9, 2016

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

#### Explanation:

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

take LHS

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

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

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

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

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

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

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

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

cancelling a $\left(1 + \tan \left(x\right)\right)$ bracket

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

as required.