# Thread: Is tan(x + 90) = -cot(x)?

1. ## Is tan(x + 90) = -cot(x)?

Is tan(x + 90) = -cot(x) ?

2. ## Re: Is tan(x + 90) = -cot(x)?

Yes, $\displaystyle tan(x+ 90)= \frac{sin(x+ 90)}{cos(x+ 90)}= \frac{cos(x)}{-sin(x)}= -cot(x)$. (I am assuming that your "x" and "90" are in degrees since that is the only way this would be true.)

3. ## Re: Is tan(x + 90) = -cot(x)?

By $\displaystyle \tan(x+90)$, I assume you mean $\displaystyle \tan(x+90^\circ)$. Use the sum of angles formulas for sine and cosine.

$\displaystyle \tan(x+90^\circ) = \dfrac{\sin(x+90^\circ)}{\cos(x+90^\circ)}$

4. ## Re: Is tan(x + 90) = -cot(x)?

$\displaystyle tan(x+90) = -cot(x)$

$\displaystyle \frac{tan(x) + tan(90)}{1-tan(x)tan(90)} = -\frac{1}{tan(x)}$

$\displaystyle tan^2(x) + tan(x)tan(90) = tan(x)tan(90) - 1$

$\displaystyle tan(x) = \sqrt{-1}$

Then what did I do wrong?

5. ## Re: Is tan(x + 90) = -cot(x)?

$\displaystyle \tan(90^\circ)$ is not defined. That is why you were told to use the sum of angles formulas for sine and cosine, not the sum of angles formula for tangent.