$\tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-\tan ^{-1}\left(\frac{\mathrm{x}-\mathrm{y}}{\mathrm{x}+\mathrm{y}}\right)=$
Answer & explanation
Correct answer: option 3
$tan^{-1}(\frac{x}{y})-tan^{-1}(\frac{\frac{x}{y}-1}{1+\frac{x}{y}})$ → Using property ⇒ $tan^{-1}a-tan^{-1}b=tan^{-1}(\frac{a-b}{1+ab})$
$tan^{-1}(\frac{x}{y})-tan^{-1}(\frac{x}{y})+tan^{-1}(1)⇒45°$
$tan^{-1}(1)=\frac{\pi}{4}$