One Last Trig Identity Q: (tanx/(secx +1)) = (secx -1/tanx)

[laura.edan]

(tanx/(secx +1)) = (secx -1/tanx)
i am completely stuck on this one...
all i got is:
L.S. = (sinx/cosx)/(1/cosx)+1
please help!
thank you

 

[skeeter]

Re: One Last Trig Identity Quesiton

start by multiplying the left side by

\(\displaystyle \frac{\sec{x}-1}{\sec{x}-1}\)

you'll also be using a form of this Pythagorean identity in your proof ...

\(\displaystyle \tan^2{x} + 1 = \sec^2{x}\)

 

[laura.edan]

Re: One Last Trig Identity Quesiton

so would that turn out to be
tanxsecx - 1 /sec^2x -1

 

[skeeter]

Re: One Last Trig Identity Quesiton

leave the numerator as \(\displaystyle \tan{x}(\sec{x} - 1)\) ... it will make life easier for you final simplification.

now, can you simplify the denominator using the Pythagorean identity from my previous post?

 

[laura.edan]

Re: One Last Trig Identity Quesiton

yes...
so that would become

tanx(secx-1)/tan^2x

which would lead to
secx-1/ tanx

which would be solved...
thank you so much for all your help again!!

 

[teacherteacher]

Re: One Last Trig Identity Quesiton

It is often helpful to keep these things in mind when proving identities:

* keep your eye on the goal
* perform the indicated operations and see what happens (as in your last question where you only needed to perform the distribution)
* try changing everything to sinx and cosx (often works)
* if there is a sum or difference in the denominator, multiply by the conjugate over itself, which is equivalent to multiplying by 1 (The conjugate of a + b is a - b and the conjugate of a - b is a + b) [This one applies to this most recent question of yours]