# Question 93fd2

Oct 9, 2016

We will use the following trigonometric identities:

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

as well as the following special products:

• ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$
• ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

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

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

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

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

=(cos(x)+sin(x))^2/((cos(x)-sin(x))(cos(x)+sin(x))#

$= \frac{\cos \left(x\right) + \sin \left(x\right)}{\cos \left(x\right) - \sin \left(x\right)}$

$= \frac{\frac{\cos \left(x\right) + \sin \left(x\right)}{\cos} \left(x\right)}{\frac{\cos \left(x\right) - \sin \left(x\right)}{\cos} \left(x\right)}$

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

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

Oct 9, 2016

$L H S = \sec 2 x + \tan 2 x$

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

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

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

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

$= \frac{1 + \tan x}{1 - \tan x} = R H S$

proved