📘 Question
If
\[
A = \frac{1}{\pi}
\begin{bmatrix}
\sin^{-1}(\pi x) & \tan^{-1}(1) \\
\sin^{-1}\left(\frac{x}{\pi}\right) & \cot^{-1}(\pi x)
\end{bmatrix}
\]
\[
B = \frac{1}{\pi}
\begin{bmatrix}
-\cot^{-1}(\pi x) & \tan^{-1}\left(\frac{x}{\pi}\right) \\
\sin^{-1}\left(\frac{x}{\pi}\right) & -\tan^{-1}(\pi x)
\end{bmatrix}
\]
Find \(A – B\).
✏️ Step-by-Step Solution
Step 1: Subtract matrices
\[
A – B = \frac{1}{\pi}
\begin{bmatrix}
\sin^{-1}(\pi x) + \cot^{-1}(\pi x) &
\tan^{-1}(1) – \tan^{-1}\left(\frac{x}{\pi}\right) \\
0 &
\cot^{-1}(\pi x) + \tan^{-1}(\pi x)
\end{bmatrix}
\]
—
Step 2: Use identities
\[
\tan^{-1}t + \cot^{-1}t = \frac{\pi}{2}, \quad \tan^{-1}(1)=\frac{\pi}{4}
\]
—
Step 3: Simplify
\[
A – B =
\frac{1}{\pi}
\begin{bmatrix}
\frac{\pi}{2} & 0 \\
0 & \frac{\pi}{2}
\end{bmatrix}
\]
—
Step 4: Final result
\[
A – B =
\begin{bmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{bmatrix}
= \frac{1}{2}I
\]
—
✅ Final Answer
\[
\boxed{(d)\; \frac{1}{2}I}
\]
—
💡 Key Concept
- \(\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}\)
- Use standard values like \(\tan^{-1}(1)=\frac{\pi}{4}\)
- Result becomes scalar multiple of identity