Mathematics - Definite Integration Question with Solution | TestHub

Given that

$${\int }_1^y{\sec }^{-1}xdx=\lambda \text{ then }{\int }_{-y}^{-1}{\sec }^{-1}x-{\tan }^{-1}\left(\sqrt{x^2-1}\right)dx+{\int }_1^y{\sec }^{-1}x-{\tan }^{-1}\left(\sqrt{x^2-1}\right)dx=\dots (|y|\ge 1)$$

equals to $\pi (y-a)-b\lambda$ then $a+b=$ $\_\_\_\_$