1. ## [SOLVED] Trig problem help please?

find the value of B given that:

Tan(3B - 4) = 1/Cot(5B - 8)

i do not know the steps to do this problem. i've tried to multiply tan and cot together but the answer wasn't correct

2. Originally Posted by something3k
find the value of B given that:

Tan(3B - 4) = 1/Cot(5B - 8)

i do not know the steps to do this problem. i've tried to multiply tan and cot together but the answer wasn't correct
hint: $\displaystyle \frac 1{\cot x} = \tan x$

and note that the period of tan(x) is $\displaystyle \pi$

3. Originally Posted by Jhevon
hint: $\displaystyle \frac 1{\cot x} = \tan x$

and note that the period of tan(x) is $\displaystyle \pi$

i am sorry i do not understand what you are trying to tell me o_o can you please explain because i am stuck at the hint

4. Originally Posted by something3k
i am sorry i do not understand what you are trying to tell me o_o can you please explain because i am stuck at the hint
what you have is $\displaystyle \tan (3B - 4) = \tan (5B - 8)$

now what does B have to be? (and remember tangent is periodic, there are infinitely many answers!)

5. Hello, something3k!

Theorem: .If $\displaystyle \tan A = \tan B$, then $\displaystyle A \:=\:B + \pi n$ .for some integer $\displaystyle n.$

Solve for $\displaystyle \theta\!:\;\;\tan(3\theta - 4) \:=\: \frac{1}{\cot(5\theta - 8)}$

We have: .$\displaystyle \tan(3\theta - 4) \;=\;\tan(5\theta - 8)$

Then: .$\displaystyle 3\theta - 4 \;=\;5\theta-8 + \pi n$

. . . $\displaystyle 2\theta \;=\;4 + \pi n \quad\Rightarrow\quad \boxed{\theta \:=\:2 + \frac{\pi}{2}n}$

6. Originally Posted by Jhevon
what you have is $\displaystyle \tan (3B - 4) = \tan (5B - 8)$

now what does B have to be? (and remember tangent is periodic, there are infinitely many answers!)

OOHHHH I GET IT OMG IM SO SLOW THANKS YOUU

7. thanks you guys, i got it now thanks so much yay