$\cos \left(2 \left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} 5\right)\right) = $ . . . . . .

Let $0 < \alpha < 1$, $\beta = \frac{1}{3\alpha}$, and $\tan^{-1}(1 - \alpha) + \tan^{-1}(1 - \beta) = \frac{\pi}{4}$. Then $6(\alpha + \beta)$ is equal to:

If $\operatorname{Tan}^{-1} \frac{1}{3}+\operatorname{Tan}^{-1} \frac{1}{7}+\operatorname{Tan}^{-1} \frac{1}{13}+\ldots+\operatorname{Tan}^{-1} \frac{1}{n^2+n+1}=\operatorname{Tan}^{-1} \theta$,then $\theta=$

The value of $\tan \left[2 \tan ^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]$ is

With usual notations in $\Delta ABC$,if $C=90^{\circ}$,then $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=$

Write the function in the simplest form: $\tan ^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} \right), x \neq 0$