Considering only the principal values of inve | vedclass

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(B) Given the equation: $	an ^{-1}(x) + 	an ^{-1}(2x) = \frac{\pi}{4}$ where $x > 0$.Using the identity $	an ^{-1}(A) + 	an ^{-1}(B) = 	an ^{-1}\

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(B) Given the equation: $	an ^{-1}(x) + 	an ^{-1}(2x) = \frac{\pi}{4}$ where $x > 0$.Using the identity $	an ^{-1}(A) + 	an ^{-1}(B) = 	an ^{-1}\left(\frac{A+B}{1-AB}ight)$,we have:$	an ^{-1}\left(\frac{x + 2x}{1 - x(2x)}ight) = \frac{\pi}{4}$Taking $	an$ on both sides:$\frac{3x}{1 - 2x^2} = 	an\left(\frac{\pi}{4}ight) = 1$$\Rightarrow 3x = 1 - 2x^2$$\Rightarrow 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 we require $x > 0$,we reject the negative root:$x = \frac{-3 + \sqrt{17}}{4}$Since $\sqrt{17} > 3$,this value is positive. Thus,there is exactly $1$ positive real value of $x$.

Considering only the principal values of inverse trigonometric functions,the number of positive real values of $x$ satisfying $\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$ is:

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