# Thread: Inverse circular relations problems

1. ## 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

Sincerely,
Carlos Jesena

2. 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

Find all values of θ:
Tan (π cot θ) = cot (π tan θ)
Hint:
Spoiler:
$\cot=\frac{1}{\tan}$ and $\arctan(x)=\text{arccot}\left(\frac{1}{x}\right)$

3. 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....

4. 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.

5. 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