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

2 Answers
Tiago Hands
Oct 20, 2016

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

#tan^2x+1=sec^2x#

Therefore:

#tan^2x=sec^2x-1#

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

#sec^2x-1=(secx+1)(secx-1)#

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

#LHS=(tan^2x)/(secx+1)+1#

#=(sec^2x-1)/(secx+1)+1#

#=((secx+1)(secx-1))/(secx+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.

#=secx-1+1#

#=secx#

#=RHS#

And here is your proof.

Related questions
See all questions in Proving Identities
Impact of this question
20497 views around the world
You can reuse this answer
Creative Commons License