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I'm trying to evaluate

$$\tan(\frac{\pi}{2}-y)$$

The answer I get is $-\cot(y)$ but the answers I have say it's $\cot(y)$

I'm not sure why it's not negative

Thanks

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  • $\begingroup$ If you show your work we may be able to point out the mistake. The correct answer is $cot(y)$; this identity is one of the "cofunction identities". $\endgroup$ – Spencer Aug 14 '16 at 5:44
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If you try and use one of those formulas with two angles in them such as, tan(u-v)=tan(u)-tan(v)/(1+tan(u)tan(v)) it won't work since tan(pi/2) is undefined.

Instead think about the relationship between sin(x) and cos(x). One of them lags or leads the other by pi/2. The function tan(x) and cot(x) follow the same principle.

The pi/2 is just a shift that makes them no longer lead or lag each other. For example cos(x-pi/2)=sin(x).

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  • $\begingroup$ Thanks a lot! That makes sense. I got the answer $\endgroup$ – Patrick Robertson Aug 14 '16 at 5:53
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Hint: Try drawing a right triangle with the other $2$ angles being $y, \frac{\pi}{2} - y$. You can then visually see that tangent of one angle is cotangent of the other. As @Spencer said if you want to know where you made a mistake please let us know what you've tried.

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