Please help me with the proof that if $\frac{1}{\cos A \cos B} + \tan A\tan B = \tan C$ then $\cos 2C \leq0$

Question

How to prove that

$$\frac{1}{\cos A \cos B} + \tan A\tan B = \tan C\qquad\Rightarrow\qquad\cos 2C \leq0.$$

Answer 1

$$\tan C=\dfrac{1+\sin A\sin B}{\cos A\cos B}$$

$$\cos2C=\dfrac{1-\tan^2C}{1+\tan^2C}=\dfrac{(\cos A\cos B)^2-(1+\sin A\sin B)^2}{(\cos A\cos B)^2+(1+\sin A\sin B)^2}$$

$$(\cos A\cos B)^2-(1+\sin A\sin B)^2=(1-\sin^2A)(1-\sin^2B)-(1+\sin A\sin B)^2$$ $$=-(\sin A-\sin B)^2\le0$$

For real $A,B$ $$(\cos A\cos B)^2+(1+\sin A\sin B)^2>0$$