# How do you prove tan^2x/(Secx+1)+1 = secx?

Apr 15, 2015

Oct 20, 2016

For reasons explained in the video below, it turns out that:

${\tan}^{2} x + 1 = {\sec}^{2} x$

Therefore:

${\tan}^{2} x = {\sec}^{2} x - 1$

Now, due to the FOIL rule (first, outer, inner, last)...

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

All of the information above combined ultimately means that...

$L H S = \frac{{\tan}^{2} x}{\sec x + 1} + 1$

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

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

*You can now get rid of (secx+1) at the top and bottom of the fraction. When the numerator and denominator of a fraction are both the same, providing they aren't both zeros, what you get is 1.

$= \sec x - 1 + 1$

$= \sec x$

$= R H S$