What is the maximum integral value of $\frac{\beta-\alpha}{\tan^{-1}\beta-\tan^{-1}\alpha}$

Question

If $0<\alpha<\beta<\sqrt3$ and if $$k=\frac{\beta-\alpha}{\tan^{-1}\beta-\tan^{-1}\alpha}$$ then find maximum value of $\lfloor k\rfloor$.

My Attempt:

Let $\beta=\tan B$ and $\alpha=\tan A$

$$k=\frac{\beta-\alpha}{\tan^{-1}\beta-\tan^{-1}\alpha}=\frac{\tan B-\tan A}{B-A}\frac{1}{\cos A\cos B}$$

$$\Longrightarrow k=\frac{\sin(A-B)}{A-B}\frac{1}{\cos A\cos B}<1\frac{1}{\cos^2 B}=\sec^2B=1+\tan^2B<1+3$$

So I get my answer as $3$.

But is it correct. Is there any way we can narrow down the values which k may take

Answer 1

Let $b=\tan(\beta),a=\tan(\alpha)$. Then $$ k = \frac{\tan(b)-\tan(a)}{b-a} $$Geometrically, this is the slope of the secant line of $y=\tan(x)$ on $[a,b]$. Tangent is convex on $[0,\arctan(\sqrt{3})=\pi/3]$, so this value is less than or equal to its derivative at $\pi/3$, which is $4$. Since we have strict inequality, we conclude the maximum $\lfloor k\rfloor $ can be is $3$; it is achieved at $\beta=1.01,\alpha=1$.