Question

- tan^2x + sec^2x = 1
Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

Answer 1

Identities are: \(\tan x=\dfrac{\sin x}{\cos x}\) and \(\sec x = \dfrac{1}{\cos x}\)

Answer 2

This "identity" doesn't seem to be true, however...
Are you sure this is the way it was stated?

Answer 3

\[-\tan^2x + \sec^2x=1\]

Answer 4

Sorry, I misread it, now it is much clearer!

Answer 5

so wouldn't
i just move -tan^2x to the right side ?

Answer 6

So after using the identities, I get:

\(\dfrac{-\sin^2x}{\cos^2x}+\dfrac{1}{\cos^2x}\)

Write it as one fraction, then use sin²x+cos²x=1

Answer 7

isnt this a trig proof?

Answer 8

I wouldn't move -tan²x to the right side, because you have to show that the left side is equal to 1.

I would just rework the left side until it's obviously equal to 1.

Answer 9

I mean this: the left side can now be written as: \(\dfrac{1-\sin^ 2x}{\cos^2x}\).

If you use that sin²x+cos²x=1, you can change the numerator to something much simpler...

Answer 10

what do i do after that

Answer 11

Well, because sin²x+cos²x=1, cos²x=1-sin²x, so you can replace the numerator with cos²x.

Answer 12

So you end up with \(\dfrac{\cos^2x}{\cos^2x}\)=...

Answer 13

thank you :)

Answer 14

yw!