Concept:tan-1x + tan-1y equals toCase:1 \(tan^{-1}(\frac{x+y}{1-xy})\) if xy < 1Case:2 \(?\ +\ tan^{-1}(\frac{x + y}{1-xy} )\) if x, y > 0 and xy > 1Case:3 \(-?\ +\ tan^{-1}(\frac{x + y}{1-xy} )\) if x, y < 0 and xy > 1Calculation:Here, \(\frac{1}{2}\times \frac{x}{3}\) < 1 as 0 < x < 6Hence, we can follow the case 1? \(\tan^{-1} \left(\frac{1}{2}\right)+\tan^{-1} \left(\frac{x}{3}\right)=\frac{\pi}{4}\)? \(tan^{-1}(\frac{\frac{1}{2}+\frac{x}{3}}{1-\frac{1}{2}\times \frac{x}{3}}) = \frac{\pi}{4}\)? \((\frac{\frac{3+2x}{6}}{\frac{6 - x}{6}}) = tan(\frac{\pi}{4})\) (? tan-1x = ? ? x = tan ?) ? \(\frac{3+2x}{6-x} = 1\) (? tan\(\frac{\pi}{4}\) = 1)? 3 + 2x = 6 - x? 3x = 3 ? x = 1