If sin x + cos x = frac{1}{5}, then tan 2x is | vedclass

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(D) Given: $\sin x + \cos x = \frac{1}{5}$Squaring both sides:$(\sin x + \cos x)^2 = (\frac{1}{5})^2$$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{1}

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$\frac{24}{7}$

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(D) Given: $\sin x + \cos x = \frac{1}{5}$Squaring both sides:$(\sin x + \cos x)^2 = (\frac{1}{5})^2$$\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{1}{25}$Since $\sin^2 x + \cos^2 x = 1$ and $2 \sin x \cos x = \sin 2x$:$1 + \sin 2x = \frac{1}{25}$$\sin 2x = \frac{1}{25} - 1 = -\frac{24}{25}$Now,$\cos^2 2x = 1 - \sin^2 2x = 1 - (-\frac{24}{25})^2 = 1 - \frac{576}{625} = \frac{49}{625}$$\cos 2x = \pm \frac{7}{25}$Since $\sin 2x$ is negative,$2x$ lies in the $III$ or $IV$ quadrant. If we assume the standard identity approach:$	an 2x = \frac{\sin 2x}{\cos 2x} = \frac{-24/25}{\pm 7/25} = \mp \frac{24}{7}$.Given the options,the magnitude is $\frac{24}{7}$.

If $\sin x + \cos x = \frac{1}{5},$ then $\tan 2x$ is

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