Question

Verify the expression:
- tan^2x + sec^2x = 1

Answer 1

HINT:
\[\tan(x) = \frac{ \sin(x) }{ \cos(x) }, \sec(x) = \frac{ 1 }{ \cos(x) }\]

Answer 2

(1/cosx^(2))-(sinx/sinx)^(2)?

Answer 3

sorry i meant (1/cosx^(2))-(sinx/cosx)^(2)?

Answer 4

Combine everything into one whole fraction. It might help you see where to go from there.

Answer 5

(1-sinx^2)/(cosx^2)

Answer 6

HINT #2:
\[\sin^2(x)-1 = \cos^2(x)\]

Answer 7

(1-sin^(2)x)/(sin^(2)x-1) ??

Answer 8

You could do that too. you should see that the proof is now complete.

Answer 9

Also:
\[\frac{ \cos^2(x) }{ \cos^2(x) }\]

Answer 10

How is 1-sin^(2)x and sin^(2)x-1 the same?

Answer 11

how does 1-sin^(2)x also equal cos^(2)x?

Answer 12

Yes, they are the same. So therefore, what do you get when you have two of the same??

AND Because: \[\sin^2(x)+\cos^2(x) = 1\] rearranging it gives you: \[\sin^2(x)-1 = \cos^2(x)\]all I did was rearrange it using basic algebra.

Answer 13

OH! Okay so since tope and bottom both equal cos^(2)x, it's 1! Thank you so much!

Answer 14

top*

Answer 15

well, you changed the denomiator to match the top so yeah.
but i was thinking of changing the numerator to the denominator. but either way works and gives you 1 = 1
hence, the proof holds true.

Answer 16

Okay cool! thanks again!