Question

Prove Trigonometric Identity

Answer 1

hmm....

Answer 2

Yeah, um this site is always really slow for me. Hang on a sec

Answer 3

\[\frac{ \tan \theta - \sec \theta+1 }{ \tan \theta + \sec \theta-1 }=\sec \theta - \tan \theta\]

Answer 4

Using your trig identities:
\[\large 1+\tan^2 \theta =\sec^2 \theta\]
\[\large 1=\sec^2 \theta-\tan^2 \theta\]
-------------------------------------------
\[\large -1=-\sec^2 \theta +\tan^2 \theta\]
\[\large LHS=\frac{\tan \theta -\sec \theta +\sec^2 \theta -\tan^2 \theta}{\tan \theta -\sec \theta -\sec^2 \theta +\tan^2 \theta}\]
\[\large =\frac{(\sec \theta-\tan \theta)(\sec \theta +\tan \theta)-(\sec\theta-\tan \theta)}{(\tan\theta -\sec \theta)(\tan \theta+\sec \theta)+(\tan\theta -\sec \theta)}\]

Answer 5

dayumm it goes back to where we started. Okay. My only choice is just put it in terms of sine and cos and then you simplify. Unfortunately I can see no easier way of doing it.

Answer 6

easier/quicker*

Answer 7

Sorry I'm still not getting this

Answer 8

Maybe this works.
\[\frac{\tan\theta -\sec \theta +1}{(\tan \theta -\sec \theta)(\tan \theta +\sec \theta+1)}=\frac{A}{\tan \theta -\sec\theta}+\frac{B}{\tan \theta +\sec\theta+1}\]

Answer 9

Have you done partial fractions?

Answer 10

Can you post the RHS again. I can't see it on my phone

Answer 11

\[=\frac{A}{\tan \theta -\sec\theta}+\frac{B}{\tan \theta+\sec\theta+1}\]

Answer 12

Yeah I can do that but surely it's not the best way

Answer 13

I tried rigging the fraction, but don't think it works.

Answer 14

Blah! This is going nowhere

Answer 15

rationalise the denominator?

Answer 16

Which one?

Answer 17

SO FAR, I ended up with
\[\frac{2\tan \theta}{(\tan \theta+\sec \theta)^2-1}\]
I haven't finished yet.

Answer 18

WHen I rationalised the denominator.

Answer 19

by multiplying both the numerator and denominator by (tan x+sex)+1

Answer 20

secx*

Answer 21

Yeah I tried that though didn't get that far

Answer 22

Nup I think I'm getting close.

Answer 23

\[\frac{2\tan\theta}{2\tan^2 \theta+2\tan\theta\sec\theta}\]

Answer 24

\[=\frac{1}{\tan \theta +\sec \theta}\]

Answer 25

Oh yep got there and took out 2tan(theta)

Answer 26

And then the 1 at the numerator use trig identity

Answer 27

And you factorise and simplify.

Answer 28

ANd you're done!!!!@!!

Answer 29

woot! That was a fun question to do.

Answer 30

\[=\frac{\sec^2 \theta-\tan^2 \theta}{\sec \theta +\tan \theta}\]

Answer 31

You can simplify that. ANd you got your answer.

Answer 32

So the trick to this question was rationalising the denominator.

Answer 33

Yeah that was awesome! Thanks heaps!

Answer 34

No worries mate.