# Ask Uncle Colin: A Tangential Conundrum

Dear Uncle Colin,

I was asked to work out $\tan\br{\theta + \piby 2}$, but the formula failed because $\tan\br{\piby 2}$ is undefined. Is there another way?

- Lost Inna Mess, Infinite Trigonometry

Hi, LIMIT, and thanks for your message!

In fact, there are several ways to approach it!

### Basic geometry

If you think of a triangle with vertices A $(0,0)$, B $(x,0)$ and C $(x,y)$, it’s clear that ABC is a right angle and $\tan(CAB) = \frac{x}{y}$.

If you rotate the triangle by a right angle anticlockwise, point C moves to C’ $(-y,x)$. That means $\tan\br{CAB + \piby 2} = -\frac{y}{x} = -\frac{1}{\tan(CAB)}$.

### Limits

Another way is to use limits: what is the limit of $\tan\br{\theta + x}$ as $x$ gets close to $\piby 2$?

We know that the addition formula for the tangent function is $\tan(\theta + x) = \frac{\tan(\theta)+\tan(x)}{1 - \tan(\theta)\tan(x)}$. However, we can rewrite that, dividing the top and bottom by $\tan(x)$, as $\frac{\frac{\tan(\theta)}{\tan(x)} + 1}{\frac{1}{\tan(x)} - \tan(\theta)}$.

As $\tan(x)$ gets large (because it’s approaching $\piby 2$), $\frac{1}{\tan(x)}$ gets very small. Let’s say that $\frac{1}{\tan(x)} = \epsilon$, giving us $\frac{\epsilon \tan(\theta)+ 1}{\epsilon - \tan(\theta)}$. As $\epsilon$ gets close to zero, this approaches $\frac{1}{-\tan(\theta)}$, as before.

### Trigonometric identities

Perhaps the least complicated way to work this out, though, is to think about the identity $\tan(A) = \frac{\sin(A)}{\cos(A)}$.

In particular, $\tan\br{\theta + \piby 2} = \frac{\sin\br{\theta + \piby 2}}{\cos\br{\theta + \piby 2}}$.

Now, you can work out that $\sin\br{\theta+\piby 2} = \cos(\theta)$ by logic or the formula; similarly, you can determine that $\cos\br{\theta+\piby2} = -\sin(\theta)$.

That gives $\tan\br{\theta+\piby2} = \frac{\cos(\theta)}{-\sin(\theta)} = -\frac{1}{\tan(\theta)}$.

Hope that helps!

- Uncle Colin