Prove LHS = RHS

[Jasonrich]

Hi all, I've been on this question for about a half an hour now

I need to prove that [tan(90-x)]/cos²x = p + 1/p ; where tanx = p

Any help would be greatly appreciated

edit: x is actually theta, on the q.paper

 

[Deleted member 4993]

Hi all, I've been on this question for about a half an hour now

I need to prove that [tan(90-x)]/cos^2x = p + 1/p ; where tanx = p

Any help would be greatly appreciated

Since you have been at it for a while - surely you have something to show for it.

If I were to solve this problem - I would first use the identity:

tan(90° - Θ) = cot(Θ)


Please share your work with us.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

We can help - we only help after you have shown your work - or ask a specific question (e.g. "are these correct?")

 

[Jasonrich]

Okay, sorry. I got to the point where I had cotΘ / cos²Θ
Then I tried making to cotΘ / (cosΘ)(cosΘ) which is cosΘ/sinΘ * [1/(cosΘ)(cosΘ)] = 1/sinΘ*cosΘ

I don't think I'm doing the right thing

 

[Jasonrich]

And then I get stuck in a loop of changing 1 to sin²Θ + cos²Θ ​and vice versa

I also tried working backwards with p + 1/p and then I end up with tanΘ + tan(90-Θ)

 

[Jasonrich]

After like an hour I got it, thanks for the help though

 

[Deleted member 4993]

Hi all, I've been on this question for about a half an hour now

I need to prove that [tan(90-x)]/cos²x = p + 1/p ; where tanx = p

Any help would be greatly appreciated

edit: x is actually theta, on the q.paper

[tan(90-x)]/cos²x = cot(x)/cos²x = cot(x) * sec²(x) = cot(x) * [1 + tan²(x)] = cot(x) + tan(x) = p + 1/p