# Is it possible to simplify this expression $\frac{(x^4 \cos x)+(4x^3 \sin x)}{(x^4 \sin x)}$?

I am having trouble figuring out if I can simplify this expression anymore. This is the most I was able to simplify it with. Let me know if anyone has ideas please! Thanks

$$\frac{(x^4 \cos x)+(4x^3 \sin x)}{(x^4 \sin x)}$$

• It's helpful to use Mathjax when you are writing your questions, as it is a little unclear what your question means. When you write (x^4cosx), do you mean (a) $x^4 \times \cos x$ or (b) $x^{4\cos x}$?
– Joe
May 29 '20 at 20:35

$$((x^4 \cos x)+(4x^3 \sin x))/(x^4 \sin x)\\ (x^4 \cos x)/(x^4 \sin x)+(4x^3 \sin x)/(x^4 \sin x)\\ \cot x + 4/x$$
$$\frac {x^4 \cos x + 4x^3 \sin x}{x^4 \sin x} = \frac {x^4\cos x}{x^4\sin x} + \frac {4x^3 \sin x}{x^4 \sin x} = \frac {\cos x}{\sin x} + \frac 4x$$.
If you like we can replace $$\frac {\cos x }{\sin x} = \cot x$$ to get $$\cot x + \frac 4x$$ but we do not have to.
And we can replace $$\frac 1x$$ with $$x^{-1}$$ if we want to get $$\cot x + 4x^{-1}$$ but again we don't have to.
Hint: let $$a=x^4 \cos x, b=4x^3 \sin x, c=x^4 \sin x$$ ,For $$a, b \in \mathbb{R}$$ and $$c\in \mathbb{R^{*}}$$ we have: $$\displaystyle\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$$