Calculus and Geometry go Hand in Hand

The maximum possible area comes out to be $\frac{1}{k}=\frac{4}{3\sqrt{3}} \\ \Rightarrow k = \frac{3\sqrt{3}}{4}$

We use the following formula which expresses $\cos \theta$ as a product of infinite terms

$\cos \theta = \prod_{r=1}^{\infty} ( 1 - \dfrac{4\theta^{2}}{ (2r-1)^{2}\pi^{2} } )$

We'll first expand the $L.H.S$ and follow it up by taking Logarithm on both sides ,

$log \cos \theta = log (1 - \dfrac{4\theta^{2}}{ \pi^{2} }) + log(1 - \dfrac{4\theta^{2}}{ 3\pi^{2} }) + log(1 - \dfrac{4\theta^{2}}{ 5\pi^{2} }) + \dots$ .

Now differentiate the above expression , to get

$\tan \theta = \dfrac{8\cdot k}{ \pi^{2} - (2 k)^{2}} + \dfrac{8\cdot k}{ 3^{2}\pi^{2} - (2 k)^{2}} + \dfrac{8\cdot k}{ 5^{2}\pi^{2} - (2 k)^{2}} + \dots$

Therefore , the sum that was asked is actually the explicit form of what we have just got .

The final answer is $\tan \frac{3\sqrt{3}}{4}^{c} = 3.5887$

Where $c$ denote Radians .