If A=1πsin-1πxtan-1xπsin-1xπcot-1πx, B=1π-cos-1πxtan-1xπsin-1xπ-tan-1πx, then A − B is equal to(a) I(b) 0(c) 2I(d) 12IDisclaimer: There is a misprint in the question. Cos−1 should be written instead of Cot−1.

If A=1πsin-1πxtan-1xπsin-1xπcot-1πx, B=1π-cos-1πxtan-1xπsin-1x

1.	If A=1πsin-1πxtan-1xπsin-1xπcot-1πx, B=1π-cos-1πxtan-1xπsin-1xπ-tan-1πx, then A − B is equal to(a) I(b) 0(c) 2I(d) 12IDisclaimer: There is a misprint in the question. Cos−1 should be written instead of Cot−1.
Answer» If $A = \frac{1}{π} [\begin{matrix} \sin^{- 1} (πx) & \tan^{- 1} (\frac{x}{π}) \\ \sin^{- 1} (\frac{x}{π}) & \cot^{- 1} (πx) \end{matrix}], B = \frac{1}{π} [\begin{matrix} - \cos^{- 1} (π x) & \tan^{- 1} (\frac{x}{π}) \\ \sin^{- 1} (\frac{x}{π}) & - \tan^{- 1} (π x) \end{matrix}]$ , then A − B is equal to (a) I (b) 0 (c) 2I (d) $\frac{1}{2} I$ Disclaimer: There is a misprint in the question. Cos⁻¹ should be written instead of Cot⁻¹.