If A0, B0 and A+B=frac{pi}{3},then the maxi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given,$A+B=\frac{\pi}{3}$.Let $y = 	an A 	an B$.Since $B = \frac{\pi}{3} - A$,we have $y = 	an A 	an(\frac{\pi}{3} - A)$.Using the $AM$-$GM$ i

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Given,$A+B=\frac{\pi}{3}$.Let $y = 	an A 	an B$.Since $B = \frac{\pi}{3} - A$,we have $y = 	an A 	an(\frac{\pi}{3} - A)$.Using the $AM$-$GM$ inequality for positive values $	an A$ and $	an B$:$\frac{	an A + 	an B}{2} \geq \sqrt{	an A 	an B}$.We know that $	an A + 	an B = \frac{\sin(A+B)}{\cos A \cos B} = \frac{\sin(\pi/3)}{\cos A \cos B} = \frac{\sqrt{3}/2}{\cos A \cos B}$.Alternatively,using the identity $	an A 	an B = \frac{\cos(A-B) - \cos(A+B)}{\cos(A-B) + \cos(A+B)}$.For a fixed sum $A+B = \frac{\pi}{3}$,the product $	an A 	an B$ is maximized when $A=B$.Thus,$A = B = \frac{\pi}{6}$.Maximum value $= 	an(\frac{\pi}{6}) 	an(\frac{\pi}{6}) = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.

If $A>0, B>0$ and $A+B=\frac{\pi}{3}$,then the maximum value of $\tan A \tan B$ is

Similar Questions

The maximum value of $12 \sin \theta - 9 \sin^2 \theta$ is

If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$,then $\tan \alpha$ equals

If $A+B+C=\pi$ and $\cos B=\cos A \cos C$,then $\tan A \tan C=$

If $A+B=\frac{\pi}{2}$,then the maximum value of $\cos A \cdot \cos B$ is

If $A, B, C$ are acute positive angles such that $A + B + C = \pi$ and $\cot A \cot B \cot C = K$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module