Let a=tanA, b=tanB, c=tanC, then sinA=a/√(1+a²), sinC=c/√(1+c²).
tan(A-B)=(a-b)/(1+ab)=(a-b)/(1+c²); (sinA/sinC)²=a²(1+c²)/(c²(1+a²)).
If we substitute b=c²/a, because tanAtanB=tan²C (assumed), then if:
tan(A-B)/tanA+(sinA/sinC)²=1⇒tanAtanB=tan²c, this substitution should produce an identity.
Thus:
(a-c²/a)/(a+ac²)+a²(1+c²)/(c²(1+a²))≡1.
Therefore:
(a-c²/a)(c²(1+a²))+a²(1+c²)(a+ac²)≡(a+ac²)(c²(1+a²)),
ac²+a³c²-c⁴/a-ac²+a³+2a³c²+a³c⁴≡ac²+a³c²+ac⁴+a³c⁴,
-c⁴-a²c²+a⁴+2a⁴c²≡a²c⁴,
a⁴(1+2c²)-a²(c²+c⁴)-c⁴≡0,
a²=c²(1+c²±√(5+10c²+c⁴))/(2(1+2c²)) is the solution of the quadratic in a².
a=±c√[(1+c²+√(5+10c²+c⁴))/(2(1+2c²))] (the negative root in parentheses is invalid).
b=c²/a. These are specific values for a, b and c. So this is not an identity:
if tan(A-B)/tanA+(sinA/sinC)²=1, then it does not follow that tanAtanB=tan²C.
FURTHER PROOF OF DISPROOF
Let A=40º, B=45º and C=37.7117º, then:
tan(A-B)/tanA+(sinA/sinC)²=1 approx.
tan²C=0.5979 approx.
tanAtanB=0.8391, so tanAtanB≠tan²C.