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

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(B) Given the equation: $	an ^{-1}(x+1)+	an ^{-1}(x-1)+	an ^{-1} x=	an ^{-1} 3$. Using the identity $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \lef

Vedclass · Accepted Answer

$1767$

Vedclass · Answer

(B) Given the equation: $	an ^{-1}(x+1)+	an ^{-1}(x-1)+	an ^{-1} x=	an ^{-1} 3$. Using the identity $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-AB} ight)$,we combine the first two terms: $	an ^{-1} \left( \frac{(x+1)+(x-1)}{1-(x+1)(x-1)} ight) + 	an ^{-1} x = 	an ^{-1} 3$. $	an ^{-1} \left( \frac{2x}{1-(x^2-1)} ight) + 	an ^{-1} x = 	an ^{-1} 3$. $	an ^{-1} \left( \frac{2x}{2-x^2} ight) + 	an ^{-1} x = 	an ^{-1} 3$. Applying the identity again: $	an ^{-1} \left( \frac{\frac{2x}{2-x^2} + x}{1 - \frac{2x^2}{2-x^2}} ight) = 	an ^{-1} 3$. $\frac{2x + 2x - x^3}{2-x^2-2x^2} = 3$. $\frac{4x-x^3}{2-3x^2} = 3$. $4x - x^3 = 6 - 9x^2$. $x^3 - 9x^2 - 4x + 6 = 0$. Since $x Dividing by $(x+1)$,we get $(x+1)(x^2 - 10x + 6) = 0$. The roots are $x = -1$ and $x = 5 \pm \sqrt{19}$. Since $x Substituting $x = -1$ into $500x^4 + 270x^2 + 997$: $500(-1)^4 + 270(-1)^2 + 997 = 500 + 270 + 997 = 1767$.

If $\tan ^{-1}(x+1)+\tan ^{-1} x+\tan ^{-1}(x-1)=\tan ^{-1} 3$,then for $x < 0$ the value of $500 x^4+270 x^2+997=$

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