The real value of x that satisfies the equati | vedclass

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(D) Given the equation: $	an ^{-1} x+	an ^{-1} 2 x=\frac{\pi}{4}$ Using the identity $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-A

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$\frac{\sqrt{17}-3}{4}$

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(D) Given the equation: $	an ^{-1} x+	an ^{-1} 2 x=\frac{\pi}{4}$ Using the identity $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-AB} ight)$,we get: $	an ^{-1} \left( \frac{x+2x}{1-x(2x)} ight) = \frac{\pi}{4}$ $\frac{3x}{1-2x^2} = 	an \left( \frac{\pi}{4} ight)$ $\frac{3x}{1-2x^2} = 1$ $3x = 1 - 2x^2$ $2x^2 + 3x - 1 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$ Since $x$ must be positive for $	an ^{-1} x + 	an ^{-1} 2x$ to be $\frac{\pi}{4}$ (as $	an ^{-1} x$ is an increasing function),we reject the negative root. Thus,$x = \frac{\sqrt{17}-3}{4}$.

The real value of $x$ that satisfies the equation $\tan ^{-1} x+\tan ^{-1} 2 x=\frac{\pi}{4}$ is

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