The value of x such that sin left(2 tan ^{-1} | vedclass

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(C) Given the equation: $\sin \left(2 	an ^{-1} \frac{3}{4}ight)=\cos \left(2 	an ^{-1} xight)$ Using the formula $\sin(2 	an^{-1} 	heta) = \f

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$\frac{1}{7}$

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(C) Given the equation: $\sin \left(2 	an ^{-1} \frac{3}{4}ight)=\cos \left(2 	an ^{-1} xight)$ Using the formula $\sin(2 	an^{-1} 	heta) = \frac{2	heta}{1+	heta^2}$ and $\cos(2 	an^{-1} x) = \frac{1-x^2}{1+x^2}$: $\frac{2 	imes \frac{3}{4}}{1+\left(\frac{3}{4}ight)^2} = \frac{1-x^2}{1+x^2}$ $\frac{\frac{3}{2}}{1+\frac{9}{16}} = \frac{1-x^2}{1+x^2}$ $\frac{\frac{3}{2}}{\frac{25}{16}} = \frac{1-x^2}{1+x^2}$ $\frac{3}{2} 	imes \frac{16}{25} = \frac{24}{25} = \frac{1-x^2}{1+x^2}$ $24(1+x^2) = 25(1-x^2)$ $24 + 24x^2 = 25 - 25x^2$ $49x^2 = 1$ $x^2 = \frac{1}{49}$ $x = \frac{1}{7}$

The value of $x$ such that $\sin \left(2 \tan ^{-1} \frac{3}{4}\right)=\cos \left(2 \tan ^{-1} x\right)$ is

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