Solve for `x :` `tan^(-1)(x/2)+tan^(-1)(x/3)=pi/4` , `0

Answer

` x=1` `quad[ 0 < x < sqrt{6}]`

Explanation

As we know that


`\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}(\frac{x+y}{1-x y})`


`\therefore \tan ^{-1} \frac{x}{2}+\tan ^{-1} \frac{x}{3}=\frac{\pi}{4}`,


`\Rightarrow \tan ^{-1}(\frac{\frac{x}{2}+\frac{x}{3}}{1-\frac{x}{2} \times \frac{x}{3}})=\frac{\pi}{4}`


`\Rightarrow \tan ^{-1}(\frac{\frac{5 x}{6}}{\frac{6-x^{2}}{6}})=\frac{\pi}{4}`


`\Rightarrow \frac{5 x}{6-x^{2}}=\tan \frac{\pi}{4}`


`\Rightarrow \frac{5 x}{6-x^{2}}=1`


`\Rightarrow 5 x=6-x^{2}`


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