Correct Answer - B
Given `sin^(-1) x + tan^(1) x = 2k + 1`
Both `sin^(-1) and tan^(-1)` are increasing functions. So `f(x)` is increasing function
The range of the function `sin^(-1) x + tan^(-1) x " is " [f(-1), f(1)]`
or `[(-3pi)/(4), (3pi)/(4)]`
Therefore, the integral values of `k` are `-1 adn 0`