Question

Verify this trigonometric equation help please!
-tan^2x+sec^2x=1

Answer 1

well you can do this..... tan^2x = (sin^2x/cos^2x) and remember that sec^2x = 1/cos^2x. just sub and do some algebra.. give it a try @jtschellenberg

Answer 2

ok i did and now i'm at (sin^2x+1)/cos^2x what next?

Answer 3

@mebs

Answer 4

It should be (-sin^2x+1)/cos^2x instead of (sin^2x+1)/cos^2x

Answer 5

wait no can someone explain?

Answer 6

so here it is -sin^2x/cos^2x + 1/cos^2x = 1

Answer 7

it has common denominators so

-sin^2x +1/cos^2x =1

Answer 8

right. what next?

Answer 9

\[LS\]\[=-tan^2x+sec^2x\]\[=-\frac{sin^2x}{cos^2x}+\frac{1}{cos^2x}\]\[=\frac{-sin^2x+1}{cos^2x}\]For the numerator, use the identity \(sin^2x+cos^2x=1\)

Answer 10

now recall that sin^2x +cos^2x = 1 right

Answer 11

yeah i remember that one

Answer 12

step by step answers posted here

http://www.mathskey.com/question2answer/4112/verify-this-trigonometric-equation-help-please

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Answer 13

just substitute -sin^2x +1 by cos^2x and wala they cancel and you get 1

Answer 14

oh so -sin^2x+1=cos^2x?

Answer 15

yes