$\tan x = -3,\,\sin x = \frac{3}{\sqrt{10}} ,\,\cos x = \frac{-1}{\sqrt{10}}$
If you work it out, $\tan 2x = \frac{3}{4}$
I don't understand how $\tan x$ is negative, but $\tan 2x$ is positive.
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Please, try to use MathJax for your typing. – Sebastián V. Romero Jan 08 '22 at 00:52
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3$2x$ is not in the 2nd quadrant, and $\tan 2x$ is not always positive when $x$ is in the 2nd quadrant, so it is not clear what the question is asking. – dxiv Jan 08 '22 at 00:56
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Just draw the graph. – TonyK Jan 08 '22 at 01:06
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The tangent function is positive in quadrants $1$ and $3$, and negative in quadrant $2$ and $4$ (you can get this by considering where sine and cosine are positive and negative, and then use quotient identities).
For your problem, $x$ and $2x$ are just the INPUT angle into the tangent function. For instance, consider the angle $x=75^{\circ}$. Since this angle is in quadrant $1$, $\tan75^{\circ}$ will be positive. However, $2x=2\cdot 75^\circ=150^\circ$, which is in quadrant $2$, and so $\tan2x=\tan150^\circ$, will be negative.
User7238
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Thank you, it is clear to me now. I really appreciate your answer. And now I have to figure out how to use MathJax. – kevin forrestal Jan 08 '22 at 01:06