If {tan ^{ - 1}}frac{{x - 1}}{{x + 2}} + {tan | vedclass

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(C) Given the equation: ${	an ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {	an ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}$Using the formula ${	an ^{ -

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$\pm \sqrt{\frac{5}{2}}$

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(C) Given the equation: ${	an ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {	an ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}$Using the formula ${	an ^{ - 1}}A + {	an ^{ - 1}}B = {	an ^{ - 1}}\left( \frac{A+B}{1-AB} ight)$,we get:${	an ^{ - 1}}\left[ \frac{{\frac{{x - 1}}{{x + 2}} + \frac{{x + 1}}{{x + 2}}}}{{1 - \left( {\frac{{x - 1}}{{x + 2}}} ight)\left( {\frac{{x + 1}}{{x + 2}}} ight)}} ight] = \frac{\pi }{4}$Simplifying the numerator and denominator:${	an ^{ - 1}}\left[ \frac{{\frac{{2x}}{{x + 2}}}}{{\frac{{{{(x + 2)}^2} - ({x^2} - 1)}}{{{{(x + 2)}^2}}}}} ight] = \frac{\pi }{4}$${	an ^{ - 1}}\left[ \frac{{2x(x + 2)}}{{{x^2} + 4x + 4 - {x^2} + 1}} ight] = \frac{\pi }{4}$${	an ^{ - 1}}\left[ \frac{{2x(x + 2)}}{{4x + 5}} ight] = \frac{\pi }{4}$Taking tangent on both sides:$\frac{{2{x^2} + 4x}}{{4x + 5}} = 	an \frac{\pi }{4} = 1$$2{x^2} + 4x = 4x + 5$$2{x^2} = 5$${x^2} = \frac{5}{2}$$x = \pm \sqrt{\frac{5}{2}}$

If ${\tan ^{ - 1}}\frac{{x - 1}}{{x + 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x + 2}} = \frac{\pi }{4}$,then $x =$

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