The number of values of  $a\in R$  for which the equation   $5{\tan }^{-1}(x^2+x+a)+3{\cot }^{-1}(x^2+x+a)=2\pi$ has unique solution, is

The given equation is 

$5{\tan }^{-1}(x^2+x+a)+3{\cot }^{-1}(x^2+x+a)=2\pi$

$\begin{array}{rcl}2{\tan }^{-1}(x^2+x+a)& =& 2\pi -\frac{3\mathrm{\pi }}{2}\\ 2{\tan }^{-1}(x^2+x+a)& =& \frac{\mathrm{\pi }}{2}\\ {\tan }^{-1}(x^2+x+a)& =& \frac{\mathrm{\pi }}{4}\end{array}$

It implies that 

$x^2+x+a=1$

Given that it has unique solution 

it means, the discriminant is zero

$\begin{array}{rcl}1-4\left(a-1\right)& =& 0\\ 4\left(a-1\right)& =& 1\\ a-1& =& \frac{1}{4}\\ a& =& \frac{5}{4}\end{array}$

Therefore, the number of vlaues of $a$ is $\boxed{\mathbf{1}}$

$2\pi =5{\tan }^{-1}(x^2+x+a)+3{\cot }^{-1}(x^2+x+a)=\frac{3\pi }{2}+2{\tan }^{-1}(x^2+x+a)$$\Rightarrow x^2+x+a=1.Forrequiredcondition\Delta =1-4(a-1)=0\Rightarrow a=\frac{5}{4}$