If $A0, B0$ and $A+B=\frac{\pi}{3}$, then the maximum value of $ an A \cdot an B$ is $\frac{1}{

Question

If $A>0, B>0$ and $A+B=\frac{\pi}{3}$, then the maximum value of $	an A \cdot 	an B$ is $\frac{1}{n}, n$ is ?

Professor PC Thomas Classes · Answer

Given $A+B=\frac{\pi}{3}$ Let $f(A)=	an A \cdot 	an B=	an A \cdot 	an \left(\frac{\pi}{3}-Aight)$\ $f^{\prime}(A)=	an A \cdot \sec ^{2}\left(\frac{\pi}{3}-Aight)(-1)$ $$ +\sec ^{2} A 	an \left(\frac{\pi}{3}-Aight)=0 $$ (For maximum or minimum of $A$ )\ $\Rightarrow A=\frac{\pi}{6}$, hence $B=\frac{\pi}{6}$\ $	herefore \quad$ Maximum value of $	an A 	an B$\ $=	an \frac{\pi}{6} \cdot 	an \frac{\pi}{6}=\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}=\frac{1}{3}$.\