Triangle Solving – Problem 6: Find the range of Which option is correct

In an acute $\triangle ABC$, let the sides opposite $A,B,C$ be $a,b,c$. Suppose $c^{2}+bc-a^{2}=0$. Find the range of $$4(\sin C+\cos C)^{2}+\frac{1}{\tan C}-\frac{1}{\tan A}.$$ Which option is correct?

A. $(4\sqrt{2},9)$ B. $(8,9)$ C. $(\frac{8\sqrt{3}}{3}+4,9)$ D. $(2\sqrt{3}+4,9)$

Step 1. From $c^{2}+bc-a^{2}=0$ we have $a^{2}-c^{2}=bc$.

Step 2. By the Law of Cosines, $a^{2}=b^{2}+c^{2}-2bc\cos A$, so $a^{2}-c^{2}=b^{2}-2bc\cos A$. Hence $bc=b^{2}-2bc\cos A$, i.e. $b=c(2\cos A+1)$.

Step 3. Using the Law of Sines $b=2R\sin B$ and $c=2R\sin C$, we get $\sin B-2\sin C\cos A=\sin C$. Since $\sin B=\sin(A+C)$, this becomes $\sin(A+C)-2\sin C\cos A=\sin C$.

Step 4. Expanding $\sin(A+C)$ gives $\sin A\cos C-\cos A\sin C=\sin C$, i.e. $\sin(A-C)=\sin C$.

Step 5. Since the triangle is acute, $A-C$ and $C$ both lie in $(0,\frac{\pi}{2})$, so $A-C=C$. Hence $A=2C$.

Step 6. Now $$4(\sin C+\cos C)^{2}+\frac{1}{\tan C}-\frac{1}{\tan A} =4\bigl(1+2\sin C\cos C\bigr)+\frac{\cos C}{\sin C}-\frac{\cos A}{\sin A}.$$

Step 7. Using $A=2C$ and identities, this simplifies to $$4+4\sin A+\frac{1}{\sin A}.$$

Step 8. Let $t=\sin A$. Because the triangle is acute and $A=2C$, we have $A\in\left(\frac{\pi}{3},\frac{\pi}{2}\right)$, so $t\in\left(\frac{\sqrt{3}}{2},1\right)$.

Step 9. The expression equals $4+4t+\frac{1}{t}$. Let $f(t)=4t+\frac{1}{t}$. Then $f'(t)=4-\frac{1}{t^{2}}>0$ on $\left(\frac{\sqrt{3}}{2},1\right)$, so $f$ is increasing.

Step 10. Hence $f(t)\in\left(\frac{8\sqrt{3}}{3},5\right)$, and the original expression lies in $\left(\frac{8\sqrt{3}}{3}+4,9\right)$. Therefore the correct choice is C.

C