is tan(π/2 - x)=cot(x) because tan in the 2nd area is "negative" and also cot(-x) equals to -cot(x)?

Question

trying to figure out if that negative/positive calculation is true. and so at the end there will be two negative and the cot(x) will be positive. is that so?

thank you.

Answer 1

Cosine, cosecant, and cotangent are so named because they are cofunctions of sine, secant, and tangent respectively. That means one function of angle $\theta$ (in radians) will be its cofunction of $\theta^c:=\pi/2-\theta$, the complement of $\theta$:

$$\cos(\theta)=\sin(\theta^c)\\ \csc(\theta)=\sec(\theta^c)\\ \cot(\theta)=\tan(\theta^c)$$

The latter two identities can be derived from the first identity. In particular,

$$\cot(\theta)={\cos \theta\over\sin\theta}={\sin (\theta^c)\over\cos(\theta^c)}=\tan(\theta^c)$$

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Also note sine is an odd function and cosine is an even function (see here): $$\sin(-\theta)=-\sin\theta\\\cos(-\theta)=\cos\theta$$

In particular, it follows that cotangent is an odd function: $$\cot(-\theta)={\cos(-\theta)\over \sin(-\theta)}={\cos(\theta)\over -\sin(\theta)}=-\cot \theta.$$