$\sec^2(\tan^{-1}3) - \tan^2(\sec^{-1}3) = $

If $y = \tan^{-1}\left(\dfrac{\log\left(\dfrac{e}{x^3}\right)}{\log ex^3}\right) + \tan^{-1}\left(\dfrac{\log(e^4x^3)}{\log\left(\dfrac{e}{x^{12}}\right)}\right)$, $x \in \left(e^{-\frac{1}{3}}, e^{\frac{1}{12}}\right)$ then $\dfrac{dy}{dx}$ is equal to...

If $0 \leq x \leq 1$ and $(\sin^{-1}x)^3 + (\cos^{-1}x)^3 = a\pi^3$ then

If $\sum\limits_{n=1}^{2026}\tan^{-1}\left(\dfrac{1}{n^2+n+1}\right) = \tan^{-1}\left(1 - \dfrac{1}{x}\right)$, where $x \neq 0$, then $x = $