1. ## f(x)=Cos(x) g(x)=Csc(x)Tan(x)

Use the given relations f and g to find (a) f · g (b) f/g (f over g) (c) f∘g (d)g∘f
f(x)=Cos(x) g(x)=Csc(x)Tan(x)

can i see the steps on how to on these?

2. first you have to fine the domain for f(x),g(x).Then give a name for each (example:h=f*g,d=f/g,l= f∘g,k=g∘f)then find the domain for the new functions(this very important to do),and for the end right the type for the new functions!!ok?

3. how do you find the domain and range for them?

4. I'll do (a)...
\begin{aligned}
f \cdot g &= (cos \, x)(csc \, x)(tan \, x) \\
&= (cos \, x)\left( \frac{1}{sin \, x} \right) \left( \frac{sin \, x}{cos \, x} \right) \\
&= 1
\end{aligned}

But this is not the same as your linear function y = 1. $f \cdot g$ is going to look like a horizontal line with some "holes". The domain of f(x) = cos(x) is all real numbers, but the domain of g(x) = csc(x)tan(x) is all reals except multiples of pi/2. So the domain of $f \cdot g$ is the intersection of the two domains, or all reals except multiples of pi/2.

5. for the f(x) we see cos which is real in real numbers so domain of f is R
for g(x) it has y=css(x) wich means that the domain of that is R-{kπ},k in Z
and u=tan(x) has domain R-{kπ+π/2}k in Z
so domain for g(x) is R-{kπ+π/2,kπ},k in Z.ok now?

6. Originally Posted by titos
for the f(x) we see cos which is real in real numbers so domain of f is R
for g(x) it has y=csc(x) wich means that the domain of that isR-{kπ},k in Z
and u=tan(x) has domain R-{kπ+π/2}k in Z
so domain for g(x) is R-{kπ+-π/2,kπ},k in Z.ok now?
That's Right!(Just do no forgot the -)
Now as eumyang said, the intersection of domains of f and g is the domain of f.g