Identity: (Sinx)/(1-Cosx) = (1+Cosx)/(Sinx)

[Julie7890]

I'm stuck on this question. It's asking me to prove the following identity:

(Tanx)/(Secx-1) = (Secx+1)/(Tanx)

I've tried it a few times, and I always simplify the left side of the equation to this:

(Sinx)/(1-Cosx)

And the right side of the eq'n to this:

(1+Cosx)/(Sinx)

And then my mind goes completely blank. There's no formula coming to mind that would prove that to be true ... could anyone give me some help on this one?

Thank you beforehand

 

[Mrspi]

I'm stuck on this question. It's asking me to prove the following identity:

(Tanx)/(Secx-1) = (Secx+1)/(Tanx)

I've tried it a few times, and I always simplify the left side of the equation to this:

(Sinx)/(1-Cosx)

And the right side of the eq'n to this:

(1+Cosx)/(Sinx)

And then my mind goes completely blank. There's no formula coming to mind that would prove that to be true ... could anyone give me some help on this one?

Thank you beforehand

I must admit that I am confused...the subject line of your question does not seem to relate to the problem you posted.

But, let's look at the problem you posted:

tan x / (sec x - 1) = (sec x + 1) / tan x

I'll choose to work with the right side of the equation. Multiply top and bottom of the fraction by tan x:

tan x / (sec x - 1) = (sec x + 1)*tan x / tan x * tan x

tan x / (sec x - 1) = tan x *(sec x + 1) / tan^2 x

Now...you've got tan^2 x in the denominator on the right side. Look at your Pythagorean identities (and the derivatives thereof). You should see that tan^2 x = sec^2 x - 1 Make that substitution:

tan x / (sec x - 1) = tan x *(sec x + 1) / (sec^2 x - 1)

Ok...now the ball is in your court. Factor the denominator of the fraction on the right side of the equation...and see if you can reduce the result.

If you are still having trouble with this, please repost showing what you have tried.

 

[Julie7890]

Thanks so much!
i had no problems after taking your advice.