θ1​,θ2​with the end of wire is written as
B=4πaμ0​I​[sinθ1​+sinθ2​]

In this case :
a=sin(2θ​)θ1​=2θ​andθ2​=90​
therefore we get magnetic field due to one wire
Blower​=4πaμ0​I​⎣⎢⎢⎢⎡​xsin(2θ​)1+sin(2θ​)​⎦⎥⎥⎥⎤​
Bupper​also same
therefore Net field
Bnet​=Blower​+Bupper​
implies that2×4πμ0​I​⎣⎢⎢⎢⎡​xsin(2θ​)1+sin(2θ​)​⎦⎥⎥⎥⎤​
2πxμ0​I​⎣⎢⎢⎢⎡​sin(2θ​)1+sin(2θ​)​⎦⎥⎥⎥⎤​2πxμ0​I​⎣⎢⎢⎢⎢⎡​2sin(4θ​)cos(4θ​){sin(4θ​)+cos(4θ​)}2​⎦⎥⎥⎥⎥⎤​​
Now dividing numerator and dinominator bycos2(4θ​)
4πxμ0​I​⎣⎢⎢⎢⎢⎡​tan(4θ​){1+tan(4θ​)}2​⎦⎥⎥⎥⎥⎤​4πxμ0​I​[{1+tan(4θ​)}2]cot(4θ​)​
Given
B=Kcot(4θ​)
And Putting the value from above expression
K=4πxμ0​I​[{1+tan(4θ​)}2]