Given:
#tan(x)tan(y)=3-2sqrt2" [1]"#
#x+y = pi/4" [2]"#
Write equation [2] as y in terms of x:
#tan(x)tan(y)=3-2sqrt2" [1]"#
#y = pi/4-x" [2.1]"#
Substitute equation [2.1] into equation [1]:
#tan(x)tan(x-pi/4)=3-2sqrt2" [1.1]"#
Use the identity #tan(A-B) = (tan(A)-tan(B))/(1+tan(A)tan(B))#
#tan(x)(tan(pi/4)-tan(x))/(1+tan(pi/4)tan(x))=3-2sqrt2" [1.2]"#
Use the fact #tan(pi/4) = 1#:
#tan(x)(1-tan(x))/(1+tan(x))=3-2sqrt2" [1.3]"#
Distribute #tan(x)# and multiply both sides by the denominator:
#-tan^2(x)+tan(x)= (3-2sqrt2)(1+tan(x))" [1.4]"#
Multiply by -1:
#tan^2(x)-tan(x)= -(3-2sqrt2)(1+tan(x))" [1.5]"#
Make the equation equal 0:
#tan^2(x)-tan(x)+(3-2sqrt2)(1+tan(x)) = 0" [1.6]"#
Distribute #(3-2sqrt2)#:
#tan^2(x)-tan(x)+3-2sqrt2+(3-2sqrt2)tan(x) = 0" [1.7]"#
Combine like terms:
#tan^2(x)+(2-2sqrt2)tan(x)+3-2sqrt2 = 0" [1.8]"#
Use the quadratic formula but only the positive value because we are told that #x > 0#:
#tan(x) =(-(2-2sqrt2)+sqrt((2-2sqrt2)^2-4(3-2sqrt2)))/2#
#x = tan^-1((-(2-2sqrt2)+sqrt((2-2sqrt2)^2-4(3-2sqrt2)))/2)#
#x = pi/8#
Use equation [2.1] to find the value of y:
#y = pi/4-pi/8#
#y = pi/8#
