The sum of three numbers is 6. Thrice the third number when added to the first number gives 7. On adding three times first number to the sum of second and third number we get 12. The product of these numbers is

$$\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=$$

The value of $$\int\limits_0^1 {{{\tan }^{ - 1}}\left( {{{2x - 1} \over {1 + x - {x^2}}}} \right)dx} $$ is

$$x=\frac{1-t^2}{1+t^2}$$ and $$y=\frac{2 a t}{1+t^2}$$, then $$\frac{d y}{d x}=$$