Question

I need help proving these trig equations
sin(x)csc(x)+ 1/(1-sin^2(x))=secx

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

Answer 1

So the first one, the setup is:

\[\sin(x)\csc(x) + \frac{1}{1 - \sin^2(x)} = \sec(x)\]

Answer 2

Yes

Answer 3

Are you 100% sure about the way I setup the first one?

Answer 4

Yea im 96-98% sure it is that way but what im unsure of is how to get 1/cos or 1+tan^2x

Answer 5

If it's not true, then there's no point in trying to verify it.

Answer 6

Alright Thank you for your help :)

Answer 7

Hang on for a minute. Where are you going?

Answer 8

nowhere i guess :o?

Answer 9

Okay, I'm going to show you a couple things

Answer 10

Alright

Answer 11

Well, the explanation is more complicated than I thought.

Answer 12

alright :o

Answer 13

What was the exact wording of the original question?

Answer 14

Prove the equations 1 step at a time

Answer 15

Okay, well, basically, I would begin like this:

\[\sin(x) \dot\ \frac{1}{\sin(x)} + \frac{1}{1 - \sin^2(x)} = \sec(x)\]

Answer 16

sin(x) and csc(x) are inverses of each other so they cancel to just 1 so what you really have is:

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

Answer 17

Which can be re-written as:

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

and combined to get:
\[\frac{1 - \sin^2x + 1}{1 - \sin^2x} = \sec(x)\]

Which simplifies to just:

\[\frac{- \sin^2x}{1 - \sin^2x} = \sec(x)\]

or
\[-\frac{\sin^2x}{1 - \sin^2x} = \sec(x)\]

Answer 18

Now we can replace 1 with the pythagorean identity: \(\sin^2x + \cos^2x = 1\)
\[-\frac{\sin^2x}{\sin^2x + \cos^2x - \sin^2x} = \sec(x)\]


Which simplifes to :
\[-\frac{\sin^2x}{\cos^2x} = \sec(x)\]

or
\[-\tan^2x= \sec(x)\]

Answer 19

Now we know this is not true because there's no way that:

\[-\frac{\sin^2x}{\cos^2x} = \frac{1}{\cos(x)}\]

Answer 20

yea

Answer 21

ive gone at it with these 2 problems for 3 hours with the result of no solution being something completely out of the question to me

Answer 22

Even if you multiplied both sides by \(\cos^2x\) to get:
\(-\sin^2x = \cos(x)\)

At that point, you could graph \(\sin^2x\) and \(\cos(x)\) and visually see that they are not equal

Answer 23

yep

Answer 24

You basically do a similar process with the second one and then realize that one isn't true either.

Answer 25

Thank you :)

Answer 26

To save yourself from all the trouble, you should get a graphing calc.

Answer 27

Yea but on tests we cant use graphing calcs

Answer 28

Sorry to hear that

Answer 29

Yea You have a good afternoon or night depending on where your from :)

Answer 30

You too