Question

If $A=\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}{\sin }^{-1}\left(\mathrm{\pi x}\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& {\cot }^{-1}\left(\mathrm{\pi x}\right)\end{array}\right],B=\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}-{\cos }^{-1}\left(\mathrm{\pi }x\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& -{\tan }^{-1}\left(\mathrm{\pi }x\right)\end{array}\right]$, 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.

Answer 1

Given: $A=\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}{\sin }^{-1}\left(\mathrm{\pi x}\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& {\cot }^{-1}\left(\mathrm{\pi x}\right)\end{array}\right],B=\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}-{\cos }^{-1}\left(\mathrm{\pi }x\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& -{\tan }^{-1}\left(\mathrm{\pi }x\right)\end{array}\right]$

$$\begin{gathered} A-B=\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}{\sin }^{-1}\left(\mathrm{\pi x}\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& {\cot }^{-1}\left(\mathrm{\pi x}\right)\end{array}\right]-\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}-{\cos }^{-1}\left(\mathrm{\pi }x\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& -{\tan }^{-1}\left(\mathrm{\pi }x\right)\end{array}\right] \\ =\frac{1}{\mathrm{\pi }}\left(\left[\begin{array}{cc}{\sin }^{-1}\left(\mathrm{\pi }x\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& {\cot }^{-1}\left(\mathrm{\pi }x\right)\end{array}\right]+\left[\begin{array}{cc}{\cos }^{-1}\left(\mathrm{\pi }x\right)& -{\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ -{\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& {\tan }^{-1}\left(\mathrm{\pi }x\right)\end{array}\right]\right) \\ =\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}{\sin }^{-1}\left(\mathrm{\pi }x\right)+{\cos }^{-1}\left(\mathrm{\pi }x\right)& {\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)-{\tan }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)\\ {\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)-{\sin }^{-1}\left(\frac{x}{\mathrm{\pi }}\right)& {\cot }^{-1}\left(\mathrm{\pi }x\right)+{\tan }^{-1}\left(\mathrm{\pi }x\right)\end{array}\right] \\ =\frac{1}{\mathrm{\pi }}\left[\begin{array}{cc}\frac{\mathrm{\pi }}{2}& 0\\ 0& \frac{\mathrm{\pi }}{2}\end{array}\right]\left(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\mathrm{\pi }}{2}\mathrm{and}{\cot }^{-1}x+{\tan }^{-1}x=\frac{\mathrm{\pi }}{2}\right) \\ =\frac{1}{\mathrm{\pi }}\times \frac{\mathrm{\pi }}{2}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right] \\ =\frac{1}{2}I \end{gathered}$$

Hence, the correct option is (d).