Question

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

Answer 1

Prove using identities.

Answer 2

My problem is I don't understand how to plug the identities :l

Answer 3

Well, there are 3 pythagorean theorem identities in trig, all 3 of them are transformations of the other.

\[\sin ^{2}x + \cos ^{2}x = 1\] is the first one. The second one I get by divided everything by sin^2(x). This gives me.
\[\frac{ \sin ^{2}x }{ \sin ^{2}x }+\frac{ \cos ^{2}x }{ \sin ^{2}x }= \frac{ 1 }{ \sin ^{2}x }\]
becomes
\[1 + \cot ^{2}x = \csc ^{2}x\]
The 3rd one comes from dividing everything in the first equation by cos^2(x)
\[\frac{ \sin ^{2}x }{ \cos ^{2}x }+\frac{ \cos ^{2}x }{ \cos ^{2}x }= \frac{ 1 }{ \cos ^{2}x }\]which becomes
\[\tan ^{2}x + 1 = \sec ^{2}x\] Often times identities require you to move one of these equations around, which is what we need to do for this problem.

Answer 4

So given the 3rd form of the pythagorean identity listed, we need to use move that around to prove this identity. Now what I'm going to do is solve for tan^2(x) in the 3rd equation. This gives me:
\[\tan ^{2}x = \sec ^{2}x - 1\]Using this, I can substitute my result for tan^2(x) in your identity to get:
\[-(\sec ^{2}x - 1) +\sec ^{2}x = 1\]
Can you kinda see what i did and what will happen? :P

Answer 5

Oooh that makes sense! Thanks for your thorough step-by-step instructions! I actually understand it now haha

Answer 6

Okay, awesome xD A lot of the time you'll have to use one of the 3 above equations and do some substitutions. Glad that made sense ^_^

Answer 7

Thanks again!!

Answer 8

yep yep :3