0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ $f(x)=\tan x$ is convex on $(0,\frac {\pi} 6)$ since its second derivative $2\tan x \sec x$ is positive. Hence $f(\frac {A+B}2 ) \leq \frac {f(A)+f(B)} 2$. $\endgroup$ Dec 9 '20 at 9:06
  • $\begingroup$ Yes, @KaviRamaMurthy, I got upto that point. But don't know how to proceed further. $\endgroup$
    – aarbee
    Dec 9 '20 at 9:08
  • 1
    $\begingroup$ The minimum is $\tan (15^{0})$. $\endgroup$ Dec 9 '20 at 9:10
1
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Yes, I got upto that point but don't know how to proceed further. $\endgroup$
    – aarbee
    Dec 9 '20 at 9:09
  • $\begingroup$ @aarbee $A+B$ is given right? plug that $\endgroup$ Dec 9 '20 at 9:10
  • $\begingroup$ oh yes, thanks. $\endgroup$
    – aarbee
    Dec 9 '20 at 9:11
1
$\begingroup$

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.