Triangle Solving – Problem 18: Find the range of

In an acute $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $b^{2}-a^{2}=ac$. Find the range of $$\frac{1}{\tan A}-\frac{1}{\tan B}.$$

Step 1. From $b^{2}-a^{2}=ac$ and the Law of Cosines $b^{2}=a^{2}+c^{2}-2ac\cos B$, we get $$ac=c^{2}-2ac\cos B\Rightarrow c=a(2\cos B+1).$$

Step 2. By the Law of Sines, $\frac{c}{a}=\frac{\sin C}{\sin A}$. Thus $$\frac{\sin(A+B)}{\sin A}=\frac{\sin C}{\sin A}=2\cos B+1.$$

Step 3. Expanding $\sin(A+B)=\sin A\cos B+\cos A\sin B$ gives $$\cos B+\frac{\cos A\sin B}{\sin A}=2\cos B+1\Rightarrow \cos A\sin B=\sin A(\cos B+1).$$

Step 4. Hence $\cos A\sin B-\sin A\cos B=\sin A$, i.e. $\sin(B-A)=\sin A$.

Step 5. Since the triangle is acute, $A$ and $B-A$ are both in $(0,\frac{\pi}{2})$, so $B-A=A$, i.e. $B=2A$.

Step 6. Now $$\frac{1}{\tan A}-\frac{1}{\tan B}=\cot A-\cot B=\frac{\sin(B-A)}{\sin A\sin B}=\frac{\sin A}{\sin A\sin B}=\frac{1}{\sin B}.$$

Step 7. With $B=2A$ and acuteness, $A\in\left(\frac{\pi}{6},\frac{\pi}{4}\right)$, so $B\in\left(\frac{\pi}{3},\frac{\pi}{2}\right)$ and $\sin B\in\left(\frac{\sqrt{3}}{2},1\right)$.

Step 8. Therefore \(\frac{1}{\tan A}-\frac{1}{\tan B}=\frac{1}{\sin B}\in\left(1,\frac{2\sqrt{3}}{3}\right).

$$(1,\frac{2\sqrt{3}}{3})$$