Considering only the principal values of an inverse function, the set

$$\mathrm{A}=\left\{x \geq 0 / \tan ^{-1} x+\tan ^{-1} 6 x=\frac{\pi}{4}\right\}$$

The solution of the equation $$\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}$$ is

The value of $$\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right), |x| < \frac{1}{2}, x \neq 0$$

If $$\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$$, then the value of $$\cos ^{-1} x+\sin ^{-1} y$$ is