Question

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

Answer 1

you need the equation for the tan and sec to be able to solve this

Answer 2

I don't understand how to use the equations or identities or whatever

Answer 3

Pythagorean Identity for Tangent: \(\large\rm \color{orangered}{\tan^2x+1=\sec^2x}\)

Do you see how you can make use of that in your problem? :)\[\large\rm - \tan^2x + \color{orangered}{\sec^2x} = 1\]

Answer 4

No, I'm really confused.

Answer 5

@zepdrix

Answer 6

This is your Pythagorean Identity,\[\large\rm \color{orangered}{\tan^2x+1=\sec^2x}\]You'll just have to acecpt that it's an identity for now.
Notice, it tells us that \(\large\rm \sec^2x\) is the same thing as \(\large\rm \tan^2x+1\)

So in our problem here,\[\large\rm - \tan^2x + \color{orangered}{\sec^2x} = 1\]we can make the exchange,\[\large\rm - \tan^2x + \color{orangered}{\tan^2x+1} = 1\]

Answer 7

And then simplify.

Answer 8

So 1=1 since -tan^2x + tan^2x cancels out, right

Answer 9

yay good job \c:/

Answer 10

What's really important in this type of problem,
is that you `only manipulate one side of the equation`.

It's not like normal algebraic stuff where we add, subtract across the equal sign.
We have to leave one side alone completely.

We didn't mess with the 1 on the right in this problem,
so we did good!

Answer 11

Thanks very much!