Is my working out for Tan's equivalent correct? So if i have:
$$\tan(\frac{\pi}{2}+\theta)$$
Am i able to:
$$\frac{\sin (\frac{\pi}{2}+\theta)}{\cos(\frac{\pi}{2}+\theta)}$$
$$=\frac {-\sin\theta}{\cos\theta} = -\tan\theta$$
Or am i misunderstanding a rule here with the fraction part? 
My aim is to show that: 
$$\tan(\frac{\pi}{2}+\theta) = -\frac{1}{\tan\theta}$$
 A: Sketch the functions for $\sin$ and $\cos$. Then hopefully you can see that $$\sin(\theta + \pi/2) = \cos\theta \ \  \hbox{ and } \ \ \cos(\theta + \pi/2) = -\sin\theta$$
This gives you the target result.
A: Your first equation is correct. Your second equation is not. For example:
$$\sin(\pi/2+0)=\sin(\pi/2)=1 \neq -\sin(0)=0.$$
What you want to use here is:
$$\sin(\pi/2+\theta)=\sin(\pi/2-(-\theta))=\cos(-\theta)$$
and the analogous identity for $\cos(\pi/2+\theta)$. These are the co-function identities; all they say is that the sine of one acute angle in a right triangle is the cosine of the other.
Alternately you can use the more general addition formulae. The case for $\sin$ looks like
$$\sin(\pi/2+\theta)=\sin(\pi/2)\cos(\theta)+\cos(\pi/2)\sin(\theta)=1 \cdot \cos(\theta) + 0 \cdot \sin(\theta) = \cos(\theta).$$
A: If you notice that $\tan\theta$ is shifted round $\frac{\pi}{2}$, then we can see that
$$ \tan\left(\frac{\pi}{2}+\theta\right) =-\cot\theta=-\frac{1}{\tan\theta}$$
There are other identities besides $\tan x=\frac{\sin x}{\cos x}$.
