Q.If $A = \dfrac{1}{\pi}\begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}\left(\dfrac{x}{\pi}\right) \\ \sin^{-1}\left(\dfrac{x}{\pi}\right) & \cot^{-1}(\pi x) \end{bmatrix}$, $B = \dfrac{1}{\pi}\begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}\left(\dfrac{x}{\pi}\right) \\ \sin^{-1}\left(\dfrac{x}{\pi}\right) & -\tan^{-1}(\pi x) \end{bmatrix}$, then $A - B$ is equal to
(A) $I$
(B) $O$
(C) $2I$
(D) $\dfrac{1}{2}I$

The scalar $\frac{1}{\pi}$ is common to both matrices, so $A-B$ equals $\frac{1}{\pi}$ times the entrywise difference. The off-diagonal terms are identical in $A$ and $B$, so they cancel; the diagonal terms combine via standard inverse-trig identities.

$A-B=\frac{1}{\pi}\begin{bmatrix}\sin^{-1}(\pi x)+\cos^{-1}(\pi x) & 0\\ 0 & \cot^{-1}(\pi x)+\tan^{-1}(\pi x)\end{bmatrix}.$ …

Setting up

Both $A$ and $B$ carry the same scalar factor $\frac{1}{\pi}$, so

and we subtract entry by entry.

The off-diagonal entries

• Row 1, column 2: $\tan^{-1}\!\frac{x}{\pi}-\tan^{-1}\!\frac{x}{\pi}=0.$
• Row 2, column 1: $\sin^{-1}\!\frac{x}{\pi}-\sin^{-1}\!\frac{x}{\pi}=0.$

These are the same in both matrices, so they vanish.

The diagonal entries

• Top-left: $\sin^{-1}(\pi x)-\big(-\cos^{-1}(\pi x)\big)=\sin^{-1}(\pi x)+\cos^{-1}(\pi x)=\frac{\pi}{2},$ using $\sin^{-1}t+\cos^{-1}t=\frac{\pi}{2}.$ …

Method: Subtracting matrices entrywise and simplifying with standard identities

For a difference $A-B$ where the entries are functions (here inverse trigonometric), factor out any common scalar, subtract corresponding entries, and simplify each entry using known identities.

Steps

Step 1: Factor out the common scalar.

If both matrices carry a factor like $\tfrac1\pi$, then $A-B=\tfrac1\pi\times(\text{entrywise difference of the inner matrices})$.

Step 2: Subtract corresponding entries.

Entries identical in $A$ and $B$ cancel to $0$ (here the off-diagonal terms), while others combine.

Step 3: Apply the relevant identities. …

Common Mistakes

Mistake 1: Not spotting the inverse-trig identities and leaving the entries unsimplified.

Why it's wrong: without $\sin^{-1}t+\cos^{-1}t=\tfrac{\pi}{2}$ and $\tan^{-1}t+\cot^{-1}t=\tfrac{\pi}{2}$, the diagonal entries cannot be reduced to $\tfrac12$. Correct approach: recognise these standard complementary identities.

Mistake 2: Mishandling the double negative when subtracting $B$'s $-\cos^{-1}(\pi x)$ entry. …