Tan3A=cot(∏/3-A), for 0≤A≤360
Here is one way. The complementary angles way.
sinA = cos(pi/2 -A)
Likewise, tanA = cot(pi/2 -A)
tan(3A) = cot(pi/3 -A)
Transform the LHS side into its cot form,
cot(pi/2 -3A) = cot(pi/3 -A)
pi/2 -3A = pi/3 -A
pi/2 - pi/3 = -A +3A
pi/6 = 2A
A = pi/12 ------------answer.
This could be an ugly problem . . .
There is a back-door approach to this problem
. . if we consider acute angles only.
If . then and are complements.
Look at this triangle:Code:* * B| * | * | y * | * A | * - - - - - - - - * x
We have: .
Since and are in the same right triangle:.
So we have: .