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$$\tan{\theta} = x$$ $\theta$ is in quadrant 2. Find values of the other 5 trig functions.

My confusion is about $x$ having to be negative because of the $-x$ value, even though it says "$x$".

$\sin{\theta} = \frac{x}{(x^2+1)^{\frac{1}{2}}}$

$\cos{\theta} = \frac{1}{(x^2+1)^{\frac{1}{2}}}$

$\tan{\theta} = x$

csc = (x^2+1)^.5/x

sec = (x^2+1)^.5

cot = 1/x

Matti P.
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anony1
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  • $\theta$ is in quadrant 2 which means $\sin$, $\csc$ are positive and $\cos$, $\sec$, $\tan$ and $\cot$ are negative. If $\tan \theta =x$, then $\sin \theta=-\frac{x}{\sqrt{x^2+1}}$. – Landuros Jul 16 '19 at 05:30
  • so everything would be the opposite of what it normally would be? – anony1 Jul 16 '19 at 05:48
  • What do you mean by "normal", anony? If $\theta$ is in quadrant 2, then $\tan\theta<0$, so $x$ is negative, and the correct formulas follow from that observation. – Gerry Myerson Jul 16 '19 at 07:02
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    Your issue is a cognitive dissonance occurring because you are thinking of $x$ as in $x$ and $y$ coordinates and the unit circle. But that is not at all how $x$ is being used here. I strongly suggest that you solve this by saying $\tan \theta = t$ instead of $x$. Then $x$ and $y$ are free to take their familiar roles in your thinking. Once you've found all the values in terms of $t$, change the variable in the expressions back from $t$ to $x$. – Paul Sinclair Jul 16 '19 at 16:44

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