determine an exact value of b such that tan(4b+3π/4)=-cot(2b-π/4) and (4b+3π/4) is in Quadrant 3. Verify your solution.how to verify this solution

tan(4b+3i/4) =
-cot(2b-pi/4)

tan(4b+3pi/4) =
-1/tan(2b-pi/4)

tan(4b+3pi/4)*tan (2b-pi/4) =
-1

tan(4b+3pi/4)* tan (2b--pi/4) + 1 =
0....(1)

But tanA*tanB +1 = tan
(A-B)/(tanA-tanB)).

Therefrore from (1), we get  tan
{(4b+3pi/4) - (2b-pi/4)}  are at right angles.

4b+3pi/4 -
(2b-pi/4) = pi/2.. (1). Or 2b-pi/4 -(4b+3pi/4) = pi/2..
(2).

2b + 3pi/4+pi/4 = pi/2.

b
=  -pi/4

4b+3pi/4 = -pi +3pi/4  = -pi+3pi/4  which is
in 4th quadrant .

If we solve (2) , -2b- pi =
pi/2.

2b+pi = -pi/2.

2b =
-3pi/2.

b = -3pi/4.

Therefore
4b+3pi/4 = -3pi + 3pi/4 = -2pi -pi+3pi/4 = -pi/4 which is 4 th
quadrant.

Therefore 4b+3pi/4 is in the 4th quadrant and not
in 3rd quadrant.