If $\mathrm{y}={\mathrm{Tan}}^{-1}\left(\frac{1}{{\cos }^{2}x+\cos x+1}\right)+{\mathrm{Tan}}^{-1}\left(\frac{1}{{\cos }^{2}x+3\cos x+3}\right)+\dots$upto n terms then $f^1\left(0\right)=$

$\mathrm{y}={\mathrm{Tan}}^{-1}\left(\frac{1}{{\cos }^{2}x+\cos x+1}\right)+{\mathrm{Tan}}^{-1}\left(\frac{1}{{\cos }^{2}x+3\cos x+3}\right)+\dots$

$y=Ta{n}^{-1}\left(\frac{\left(1+\cos x\right)-\cos x}{1+\cos x\left(1+\cos x\right)}\right)+Ta{n}^{-1}\left(\frac{\left(2+\cos x\right)-\left(1+\cos x\right)}{1+\left(2+\cos x\right)\left(1+\cos x\right)}\right)+.......$

$y=Ta{n}^{-1}\left(1+\cos x\right)-Ta{n}^{-1}\left(\cos x\right)+Ta{n}^{-1}\left(2+\cos x\right)-Ta{n}^{-1}\left(1+\cos x\right)+....$

$y=Ta{n}^{-1}\left(n+\cos x\right)-Ta{n}^{-1}\left(\cos x\right)$

$\frac{dy}{dx}=\frac{-\sin x}{1+{\left(n+\cos x\right)}^2}-\frac{-\sin x}{1+{\cos }^{2}x}$

$x=0\Rightarrow {\left(\frac{dy}{dx}\right)}_{x=0}=0$