If A>0 , B>0 and A+B=∩/3 , then the maximum value of tan A tan B is
a) 1/6
b)1/3
c)1/2
d)1
ans b)
.............solve it
1applying jenson on ln(tanx)
\frac{\partial ^2 \left( \ln {\tan x}\right)}{\partial x^2}=\sec^2x-\csc ^2 x \leq 0 \ \ \forall x\in[0,\frac{\pi}{4}] \\ \\\texttt{hence the function is concave }\\ \\ \frac{\ln {\tan A}+\ln {\tan B}}{2} \leq \ln {\tan \frac{(A+B)}{2}} \\ \ln(\tan A.\tan B)\leq \ln \frac{1}{3} \\ \tan A.\tan B \leq \frac{1}{3}
1use AM-GM inequality
tan(A)+tan(B)2 ≥ √tan(A)tan(B)
all are positive so
\left(\frac{tan(A)+tan(B)}{2} \right)^{2}\geq tan(A)tan(B)
now maximize { tan(A) + tan(B) } it comes equal to2√3
put this value in eqn and it gives us result
tan(A)tan(B)≤13
