Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

- tan2x + sec2x = 1

Answer 1

HI!!

Answer 2

Heyy!

Answer 3

is that \[-\tan^2(x)+\sec^2(x)=1\]?

Answer 4

Yes.

Answer 5

ok lets start with the mother of all trig identities
\[\sin^2(x)+\cos^2(x)=1\]

Answer 6

divide each term by \(\cos^2(x)\)

Answer 7

you get
\[\frac{\sin^2(x)}{\cos^2(x)}+\frac{\cos^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}\\
\tan^2(x)+1=\sec^2(x)\]

Answer 8

subtract \(\tan^2(x)\) from both sides and you get exactly what you want
\[-\tan^2(x)+\sec^2(x)=1\]

Answer 9

I like your smile Misty.

Answer 10

i'd walk a mile...

Answer 11

you good with that?

Answer 12

@unknownunknown thanks! i like @clairvoyant1 smile too

Answer 13

Yes, you both have a beautiful smile.

Answer 14

Yes, so
1 - sin^2(x) = cos^2(x)

Answer 15

yeah you can use that too
then divide by \(\cos^2(x)\) to get what you want

Answer 16

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

Answer 17

you got \[1-\sin^2(x)=\cos^2(x)\] right ?

Answer 18

oh yeah, what you wrote
except you forgot the right hand side of the equal sign

Answer 19

yes.

Answer 20

ok so the first term you wrote above is \(-\frac{\sin^2(x)}{\cos^2(x)}\)
that is the same as \(-\tan^2(x)\)

Answer 21

the second term you wrote is \(\frac{1}{\cos^2(x)}\) which is \(\sec^2(x)\)

Answer 22

and on the right side of the equal sign you would have \(\frac{\cos^2(x)}{\cos^2(x)}\) which is \(1\)

Answer 23

@clairvoyant1 you got that? misty wrote it all out should be good right?

Answer 24

and that's the rhs right?

Answer 25

yes the rhs is 1

Answer 26

so you are done at that step with
\[-\tan^2(x)+\sec^2(x)=1\] which is what you wanted to show