1. Find>trigo~!

Tan3A=cot(∏/3-A), for 0≤A≤360

Tan3A =????

2. Here is one way. The complementary angles way.

sinA = cos(pi/2 -A)
Likewise, tanA = cot(pi/2 -A)
So,
tan(3A) = cot(pi/3 -A)
Transform the LHS side into its cot form,
cot(pi/2 -3A) = cot(pi/3 -A)
So,
pi/2 -3A = pi/3 -A
pi/2 - pi/3 = -A +3A
pi/6 = 2A

3. Hello, sanikui!

This could be an ugly problem . . .

$\displaystyle \tan 3x\:=\:\cot(60^o -x),\;\text{ for }0^o\,\leq\,x \leq \,360^o$

$\displaystyle \text{Evaluate: }\tan3x$
There is a back-door approach to this problem
. . if we consider acute angles only.

If $\displaystyle \tan A \:=\:\cot B$. then $\displaystyle A$ and $\displaystyle B$ are complements.

Look at this triangle:
Code:
                        *
* B|
*     |
*        | y
*           |
* A            |
* - - - - - - - - *
x

We have: .$\displaystyle \tan A \:=\:\frac{y}{x},\;\;\cot B \:=\:\frac{y}{x}$

Since $\displaystyle A$ and $\displaystyle B$ are in the same right triangle:. $\displaystyle A + B \:=\:90^o$

So we have: .$\displaystyle 3x + (60^o - x) \;=\;90^o\quad\Rightarrow\quad 2x \:=\:30^o \quad\Rightarrow\quad x \:=\:15^o$

Therefore: .$\displaystyle \tan(3x) \;=\;\tan45^o \;=\;1$