Question

Steps on how to establish this identity:
(tanθ+secθ-1)/(tanθ-secθ+1) = tanθ+secθ

Answer 1

let me do on paper first

Answer 2

good question ...Gimme some more time ..Do you have any ideas?

Answer 3

That's fine! I'm still working on it... I tried rewriting the left side and I ended up with
(sinθ-cosθ+1)/sineθ+cosθ-1), but I'm stuck

Answer 4

ya I first thought of that ..But it seems there is no need for that both LHS and RHS have tan theta and cos theta ..So we dont need to write tan theta in terms of sin and cos

Answer 5

*tan theta and sec theta

Answer 6

I think I got it :)

Answer 7

YES! :D

Answer 8

u also gt it ? :)

Answer 9

Maybe, I'm still checking though

Answer 10

ok first you say ur answer then I will say mine

Answer 11

Ok, it's going to be really long, so you'll have to wait a while as I type lol

Answer 12

yep I will :)

Answer 13

...........

Answer 14

lol

Answer 15

\[\frac{\tanθ+(\secθ-1)}{\tanθ- (\secθ-1)}· \frac{\tanθ+(\secθ-1)}{\tanθ+(\secθ-1)}= \frac{ \tan²θ+2\tanθ(\secθ-1)+\sec²θ-2\secθ+1}{\tan²θ-(\sec²θ-2\secθ+1)}\]

Answer 16

wth that took forever

Answer 17

wow :) I did exactly the same

Answer 18

nah not really ..hw abt this

Answer 19

What I did

\[\dfrac{\tan \theta +(\sec \theta -1)}{\tan \theta -(\sec \theta -1) }=\dfrac{\tan \theta +(\sec \theta -1)}{\tan \theta -(\sec \theta -1) } \times \dfrac{\tan \theta +(\sec \theta -1)}{\tan \theta +(\sec \theta -1) }\]

\[=\dfrac{\tan^2 \theta +\sec ^2 \theta +1-2\sec \theta +2\tan \theta(\sec \theta -1)}{\tan^2 \theta-\sec^2 \theta +2\sec \theta -1}\]

\[=\dfrac{2\sec^2 \theta-2 \sec \theta +2\tan \theta(\sec \theta-1)}{2(\sec \theta-1)}\]

\[=\dfrac{2\sec \theta(\sec \theta-1)+2\tan \theta(\sec \theta-1)}{2(\sec \theta-1)}\]

Cancelling off \(2(\sec \theta-1)\) on numerator and denominator .

=\(\tan \theta +\sec \theta=RHS \)

Answer 20

WOW yours looks amazing! :) Better than mine! I was like okay first step and you're done lol. Typing in the equations took forever.. it amazes me how you could type that!

Answer 21

;)

Answer 22

Anyway glad that you figured it out yourself!

Answer 23

It took a while! :) Thanks so much! I appreciate your help!

Answer 24

yw :)