Finding minimum value of $\tan A+\tan B$, given $A+B=30^\circ$

If $$A>0$$, $$B>0$$ and $$A+B=30^\circ$$, find the minimum value of $$\tan A+\tan B$$.

A similar question has been asked here. Many good answers are posted there. I understood that.

One particular answer by @Bill Kleinhans intrigued me as it was short and straightforward. But I didn't fully understand that answer. It uses Jensen's inequality, which I know, but don't know how to use that to get the answer here.

Not posting this as a comment there because Bill has been away close to three years now.

• $f(x)=\tan x$ is convex on $(0,\frac {\pi} 6)$ since its second derivative $2\tan x \sec x$ is positive. Hence $f(\frac {A+B}2 ) \leq \frac {f(A)+f(B)} 2$. Dec 9 '20 at 9:06
• Yes, @KaviRamaMurthy, I got upto that point. But don't know how to proceed further. Dec 9 '20 at 9:08
• The minimum is $\tan (15^{0})$. Dec 9 '20 at 9:10

Bill's idea may be this..

consider $$f(x)=\tan x$$ $$f''(x)=2\sec^2 x \tan x>0$$ for $$x\in (0,\pi/3)$$

.By jensen $$f(x)+f(y)\le \frac{f(x)+f(y}{2}\Rightarrow \frac{\tan A+\tan B}{2}\ge {\tan(\frac{A+B}{2})}=\tan 15^o$$

• Yes, I got upto that point but don't know how to proceed further. Dec 9 '20 at 9:09
• @aarbee $A+B$ is given right? plug that Dec 9 '20 at 9:10
• oh yes, thanks. Dec 9 '20 at 9:11

$$\tan A+\tan B=\dfrac{\sin(A+B)}{\cos A\cos B}=\dfrac{2\sin(A+B)}{\cos(A-B)+\cos(A+B)} =\dfrac1{\cos(2A-30^\circ)+\dfrac{\sqrt3}2}$$

So, we need to maximize $$\cos(2A-30^\circ)+\dfrac{\sqrt3}2$$

Now $$-30^\circ<2A-30^\circ<30^\circ\implies\cos30^\circ<\cos(2A-30^\circ)\le1$$