# Inverse circular relations problems

• April 23rd 2010, 09:01 PM
CarlosJesena
Inverse circular relations problems
Hey guys, my teacher gave me these two problems which have been keeping me up all night:

Prove:
[2tan^-1 (1/3) ]+ [tan^-1(1/7)] = pi/4

and

Find all values of θ:
Tan (π cot θ) = cot (π tan θ)

My classmates and I have been kinda stumped... Any help is much appreciated http://www.mymathforum.com/images/sm...on_biggrin.gif

Sincerely,
Carlos Jesena
• April 23rd 2010, 09:44 PM
Drexel28
Quote:

Originally Posted by CarlosJesena
Hey guys, my teacher gave me these two problems which have been keeping me up all night:

[2tan^-1 (1/3) ]+ [tan^-1(1/7)] = pi/4

Hint:
Spoiler:
Take the tangent of both sides and remember this

Quote:

Find all values of θ:
Tan (π cot θ) = cot (π tan θ)
Hint:
Spoiler:
$\cot=\frac{1}{\tan}$ and $\arctan(x)=\text{arccot}\left(\frac{1}{x}\right)$
• April 23rd 2010, 09:52 PM
CarlosJesena
Quote:

Originally Posted by Drexel28
Hint:
Spoiler:
Take the tangent of both sides and remember this

Hint:
Spoiler:
$\cot=\frac{1}{\tan}$ and $\arctan(x)=\text{arccot}\left(\frac{1}{x}\right)$

Thanks for the help! For this first problem, I turned it into 2A + B = pi/4 then that becomes tan (2A+B) = 1. Then I expanded the left side and got:

[tan(2A)+tan(B)] / [1-(tan(2A)tan(B))]

Am I on the right track? I was thinking of expanding the tan 2A's into double angle formula....
• April 23rd 2010, 10:12 PM
Drexel28
Quote:

Originally Posted by CarlosJesena
Thanks for the help! For this first problem, I turned it into 2A + B = pi/4 then that becomes tan (2A+B) = 1. Then I expanded the left side and got:

[tan(2A)+tan(B)] / [1-(tan(2A)tan(B))]

Am I on the right track? I was thinking of expanding the tan 2A's into double angle formula....

Yes, you're on the right track.
• April 23rd 2010, 10:34 PM
CarlosJesena
Quote:

Originally Posted by Drexel28
Yes, you're on the right track.

Thanks so much for the help! I get number one now. Just number two is still a mystery