If $\sum_{n=1}^{2026} \tan^{-1}(\frac{1}{n^2+n+1}) = \tan^{-1}(1 - \frac{1}{x})$, where $x \neq 0$, then $x = $

If $\cos^{-1} x = \alpha$ $(0 < x < 1)$ and $\sin^{-1} (2 x \sqrt{1 - x^2}) + \sec^{-1} (\frac{1}{2 x^2 - 1}) = \frac{2 \pi}{3}$,then $\alpha$ is

$\cos ^{ - 1}\frac{4}{5} + \tan ^{ - 1}\frac{3}{5} = $

$\tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2a}}{{1 + {a^2}}}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {a^2}}}{{1 + {a^2}}}} \right)} \right] = $

If $\cos \left(\cos ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} x\right)=1$,then find the value of $x$.

If $0 \leqslant \cos ^{-1} x \leqslant \pi$ and $\frac{-\pi}{2} \leqslant \sin ^{-1} x \leqslant \frac{\pi}{2}$,then at $x=\frac{1}{5}$ the value of $\cos \left(2 \cos ^{-1} x+\sin ^{-1} x\right)$ is