IIT-JEE 1993Question 1
If $A>0,\ B>0$ and $A+B=\pi/3$, then the maximum value of $\tan A\tan B$ is ______.
Trigonometric Ratio & Identities
Answer$\dfrac13$
Solution
Concept. Write $\tan A\tan B$ with $\cos(A-B)$ and $\cos(A+B)=\frac12$; it increases with $\cos(A-B)$. Formulas used. (1) $2\sin A\sin B=\cos(A-B)-\cos(A+B)$, $2\cos A\cos B=\cos(A-B)+\cos(A+B)$Step 1 - Rewrite. By (1), with $c=\cos(A-B)$: $$\tan A\tan B=\frac{c-\frac12}{c+\frac12}=1-\frac{1}{c+\frac12}.$$ Step 2 - Maximise. This increases with $c$, and $c\le1$, with $c=1$ at $A=B=\frac\pi6$. Step 3 - Value. $\dfrac{1/2}{3/2}=\dfrac13$. Answer: $\dfrac13$