# How do you simplify the expression (cottsec^2t-cott)/(sint tant+cost)?

$\sin t$

#### Explanation:

(I'm writing this bit - the intro - last. I've just worked the problem and I had no idea where it would go or how to simplify it, beyond the fact that I took it apart piece by piece and took substitution opportunities where I could. It's easy to look at something complicated and shrug and think "I can't do this", but if you can start by doing something - anything - you can start to make sense of it. This is not to say that the first few things you do will make the problem simple right away - it can get rather messy. But stick with it and it should resolve itself into something simpler!)

$\frac{\cot t {\sec}^{2} t - \cot t}{\sin t \tan t + \cos t}$

In the numerator, let's factor out $\cot t$ to get to a $\left({\sec}^{2} t - 1\right)$ term and in the denominator let's look at multiplying the sin and tan so that we can work with the cos:

$\frac{\cot t \left({\sec}^{2} t - 1\right)}{\sin t \left(\sin \frac{t}{\cos} t\right) + \cos t}$

In the numerator, we can use the trig identity ${\tan}^{2} t = {\sec}^{2} t - 1$ and substitute in. In the denominator, I'm going to work on getting the terms to add:

$\frac{\cot t \left({\tan}^{2} t\right)}{\left({\sin}^{2} \frac{t}{\cos} t\right) + \cos t \left(\cos \frac{t}{\cos} t\right)}$

$\frac{\left(\frac{1}{\tan} t\right) \left({\tan}^{2} t\right)}{\left({\sin}^{2} \frac{t}{\cos} t\right) + \left({\cos}^{2} \frac{t}{\cos} t\right)}$

$\tan \frac{t}{\frac{{\sin}^{2} t + {\cos}^{2} t}{\cos} t}$

$\frac{\tan t \cos t}{{\sin}^{2} t + {\cos}^{2} t}$

We can substitute in the denominator, using the trig identity ${\sin}^{2} t + {\cos}^{2} t = 1$

$\frac{\left(\sin \frac{t}{\cancel{\cos}} t\right) \cancel{\cos} t}{1}$

$\sin t$