Question

verify the identity
1 + (sec^2)x *(sin^2)x = (sec^2)x

Answer 1

what's the most fundamental trig identity

Answer 2

cos^2x+sin^2x=1 so (if you divide both sides by cos^2x)...
1+tan^2x=sec^2x
tan^2x=sec^2x-1

sec^2x/sec^2x-1=csc^2x
sec^2x/tan^2x=csc^2x
(1/cos^2x)/(sin^2x/cos^2x)=csc^2x
(1/cos^2x)(cos^2x/sin^2x)=csc^2x
(1/sin^2x)=csc^2x
csc^2x=csc^2x

Answer 3

Why is
\[
(sec^2)x *(sin^2)x=\tan^2(x)
\]

Answer 4

*hint* most people are skipping this part, but i think it is crucial:

sec(x) = 1/cos(x)

sec^2(x) = 1/cos^2(x)

Answer 5

So
\[
1+ \tan^2(x)=\sec^2(x)
\]

Answer 6

All trig functions are of x, deleted for brevity.

Given cos^2 + sin^2 =1, multiply both sides by sec^2. Your conclusion follows directly.

Answer 7

everyone's saying different things :/ im confused...

Answer 8

Not really. Everything starts from sin^2 + cos^2 = 1

Answer 9

Now it's true, you have to learn to forget about x or theta what angle the sine is for.
And you also learn to forget about the radius, because this is unit circle.

Answer 10

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

Answer 11

Good work with the equation editor @eliassaab

Answer 12

thank u everyone! :)

Answer 13

That is a summary of what was posted above
\[
1 + \sec^2x \, \sin^2x =\\
1 + \frac 1{\cos^2 x} \sin^2x= \frac {\sin^2 x+\cos^2 x}{\cos^2 x}=
\frac 1{\cos^2(x)}=\sec^2(x)


\]

Answer 14

wait @ eliassaab i dont understand how you got sin2x+cos2x/cos2x

Answer 15

same denominator

Answer 16

of what?

Answer 17

Here are more details
\[
1 + \sec^2x \, \sin^2x =\\
1 + \frac 1{\cos^2 x} \sin^2x= \frac{ \cos^2 x}{\cos^2 x}+\frac{\sin^2x}{\cos^2 x}=\\
\frac {\sin^2 x+\cos^2 x}{\cos^2 x}=
\frac 1{\cos^2(x)}=\sec^2(x)


\]

Answer 18

Did you get it now?

Answer 19

where did the one go?

Answer 20

\[
1= \frac{ \cos^2 x}{\cos^2 x}
\]

Answer 21

Are you still there?

Answer 22

yes so (1/cos2x)*sin2x=sin2x/cos2x??

Answer 23

@eliassaab

Answer 24

@telliott99

Answer 25

Correct, sarah.

Answer 26

Sorry I was gone, but yes you are correct. Go back through and look carefully at @eliassaab stuff.

Answer 27

ok. thank u everyone