Given that ∫↖{y}↙{1} sec^{-1} x dx = λ and for (y ≥ 1) . If ∫↖{-1}↙{-y} {sec^{-1} x - tan^{-1}(√{x^2-1})} dx + ∫↖{y}↙{1} {sec^{-1} x - tan^{-1} (√{x^2-1})} dx = π (y - a) - bλ , then a + b = ....

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