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

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

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$1$

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(A) Given the equation: $	an^{-1}(x+2) + 	an^{-1}(x-2) = 	an^{-1}(\frac{1}{2})$Using the formula $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}(\frac{A+B}{1-AB})$,we get:$	an^{-1}(\frac{(x+2) + (x-2)}{1 - (x+2)(x-2)}) = 	an^{-1}(\frac{1}{2})$Simplifying the expression inside the $	an^{-1}$ function:$\frac{2x}{1 - (x^2 - 4)} = \frac{1}{2}$$\frac{2x}{1 - x^2 + 4} = \frac{1}{2}$$\frac{2x}{5 - x^2} = \frac{1}{2}$Cross-multiplying gives:$4x = 5 - x^2$$x^2 + 4x - 5 = 0$Factoring the quadratic equation:$(x+5)(x-1) = 0$So,$x = 1$ or $x = -5$.Checking the values:If $x = -5$,then $	an^{-1}(-3) + 	an^{-1}(-7)$,which is negative,while $	an^{-1}(\frac{1}{2})$ is positive. Thus,$x = -5$ is rejected.If $x = 1$,then $	an^{-1}(3) + 	an^{-1}(-1) = 	an^{-1}(3) - 	an^{-1}(1) = 	an^{-1}(\frac{3-1}{1+3}) = 	an^{-1}(\frac{2}{4}) = 	an^{-1}(\frac{1}{2})$. This is valid.The only valid value is $x = 1$. The sum of the values is $1$.

If $\tan^{-1} (x+ 2)+ \tan^{- 1}( x -2)= \tan^{-1} (\frac{1}{2}),$ then the sum of the value$(s)$ of $x$ is equal to-

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