# 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

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To solve this problem, we need to remember the half pi shift identity for cot:

tan(A) = -cot( A + pi/2 )

So:

A = 4b + 3pi/4

A + pi/2 = 2b - pi/4

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

2b = -pi - pi/2

b = -3pi/4, which is in the third quadrand

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.