Question

- tan2x + sec2x = 1

Answer 1

Prove?

Answer 2

yes

Answer 3

Pythagorean Identity :)
cos²x + sin²x = 1
Are you allowed to use this? ^.^

Answer 4

\[\sec2x=(1+\tan2x)\\
\sec^22x=(1+\tan2x)^2=1+\tan^22x
\]

Answer 5

Simply differentiate it....

Answer 6

Just because derivatives are equal doesn't mean two expressions are equal :)

Answer 7

yes i am allowed to use that @PeterPan

Answer 8

\[\Large f'(x) = g'(x) \ \ \rightarrow \ \ f(x)=g(x)+C\]
They differ by a constant, @goformit100 :)

@bbfoster123
Well
cos²x + sin²x = 1
See what happens when you divide everything by cos²x ^.^

Answer 9

it does not work as you see

Answer 10

Just differentiate it

Answer 11

is it proove or solve for "x"???

Answer 12

yes prove it then

Answer 13

No solve for x, @electrokid
It's an identity... any value for x in which isn't a multiply of pi would make this expression true.

Answer 14

The given statement is not true for all "x"

Answer 15

i am so confused....

Answer 16

@bbfoster123 state your question explicitly

Answer 17

I see. Well, maybe we need to clear up some confusion.
@bbfoster123
Is it
-tan(2x) + sec(2x) = 1
or
-tan²x + sec²x = 1

?

Answer 18

verify the 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 19

Yeah, but is the "2" an exponent?

Answer 20

@PeterPan the second equation u stated

Answer 21

Well then, it should be easy :)
Start with the Pythagorean identity ^.^
\[\Large \sin^2 x+\cos^2x=1\]
Okay?

Answer 22

ok i got that

Answer 23

Now, you can divide both sides by cos²x
\[\Large \frac{\sin^2x}{\cos^2x}+\frac{\cos^2x}{\cos^2x}=\frac1{\cos^2x}\]

Answer 24

And this is the same as
\[\Large \left(\frac{\sin x}{\cos x}\right)^2+1=\left(\frac1{\cos x}\right)^2\]

Answer 25

ok

Answer 26

Well? You're done :)
Well, almost ^.^
What's
\[\Large \frac{\sin x}{\cos x}\]?

Answer 27

tangent

Answer 28

And what's\[\Large \frac1{\cos x}\]?

Answer 29

sec x

Answer 30

That's right :)
So we can rephrase the equation as...
\[\huge \tan^2x+1=\sec^2x\]
And now, just rearrange as necessary :)

Answer 31

thank you so much i understand it now.

Answer 32

No problem :)