Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Squaring and adding the given equations yields $2+2\cos (A-B)=2$, so $\cos (A-B)=0$, implying $A-B=\frac{\pi }{2}$ for a triangle. Substituting $A=B+\frac{\pi }{2}$ into the original equations and dividing them gives $\frac{\cos B+\sin B}{\cos B-\sin B}=\tan C$, which simplifies to $\tan (\frac{\pi }{4}+B)=\tan C$. From $A+B+C=\pi$, we have $2B+C=\frac{\pi }{2}$. Solving these equations for $A,B,C$ gives $A=\frac{7\pi }{12}$, $B=\frac{\pi }{12}$, $C=\frac{\pi }{3}$. Now, $\tan A=-(2+\sqrt{3})$, $\tan B=2-\sqrt{3}$, and $\tan C=\sqrt{3}$. Therefore, $\tan A+\tan B=-2\sqrt{3}$, so $\frac{\tan A+\tan B}{\tan C}=-2$. The required value is $(-2{)}^2=4$.