# How do you verify 1-tan^2x=2-sec^2x?

Mar 5, 2018

Use the Pythagorean identity:

$\textcolor{w h i t e}{\implies} {\sin}^{2} x + {\cos}^{2} x = 1$

Then, divide all the terms by ${\cos}^{2} x$ to derive a new identity:

$\textcolor{w h i t e}{\implies} {\sin}^{2} \frac{x}{\cos} ^ 2 x + {\cos}^{2} \frac{x}{\cos} ^ 2 x = \frac{1}{\cos} ^ 2 x$

$\textcolor{w h i t e}{\implies} {\tan}^{2} x + 1 = {\sec}^{2} x$

Now, here's the actual proof (starting with the right side):

$R H S = 2 - \textcolor{red}{{\sec}^{2} x}$

$\textcolor{w h i t e}{R H S} = 2 - \left(\textcolor{red}{1 + {\tan}^{2} x}\right)$

$\textcolor{w h i t e}{R H S} = 2 - 1 - {\tan}^{2} x$

$\textcolor{w h i t e}{R H S} = 1 - {\tan}^{2} x$

$\textcolor{w h i t e}{R H S} = L H S$