If tan ^{-1}(x+2)+tan ^{-1}(x-2)-tan ^{-1}lef | vedclass

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(C) Given equation: $	an ^{-1}(x+2)+	an ^{-1}(x-2) = 	an ^{-1}\left(\frac{1}{2}ight)$ Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1}

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$1$

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(C) Given equation: $	an ^{-1}(x+2)+	an ^{-1}(x-2) = 	an ^{-1}\left(\frac{1}{2}ight)$ Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1}\left(\frac{A+B}{1-AB}ight)$,we get: $	an ^{-1}\left(\frac{(x+2)+(x-2)}{1-(x+2)(x-2)}ight) = 	an ^{-1}\left(\frac{1}{2}ight)$ $	an ^{-1}\left(\frac{2x}{1-(x^2-4)}ight) = 	an ^{-1}\left(\frac{1}{2}ight)$ $	an ^{-1}\left(\frac{2x}{5-x^2}ight) = 	an ^{-1}\left(\frac{1}{2}ight)$ Comparing both sides: $\frac{2x}{5-x^2} = \frac{1}{2}$ $4x = 5 - x^2$ $x^2 + 4x - 5 = 0$ $(x+5)(x-1) = 0$ Thus,$x = -5$ or $x = 1$.

If $\tan ^{-1}(x+2)+\tan ^{-1}(x-2)-\tan ^{-1}\left(\frac{1}{2}\right)=0$,then one value of $x$ is

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