# Why can't $\frac{1}{\sin(x)\cos(x)}$ be expressed in the form $\frac{A}{\sin x}+\frac{B}{\cos x}$

I've tried expressing $\frac{1}{\sin(x)\cos(x)}$ as partial fractions:

$$\frac{1}{\sin(x)\cos(x)} = \frac{A}{\sin x}+\frac{B}{\cos x} \implies 1=A\cos x + B\sin x$$

I let $x=\frac{\pi}{2}$, getting $A=-1$. Then I let $x=\pi$, getting $B=1$. This means that $1=\cos x-\sin x$ which is obviously wrong most of the time.

But why is it wrong? If the denominator had something like $(x^2-1)$ I still can get a correct answer for different values I substitute.

• Algebraically speaking, partial fractions only works on integral domain. Analytically speaking, trig functions are different from polynomial functions. If you really want to decompose, then $A,B$ are not necessarily real numbers anymore. So you may have $$\frac 1{\sin(x)\cos(x)} = \tan(x) + \cot(x)$$ by taking $A = \cos(x), B = \sin(x)$.
– xbh
Sep 12, 2018 at 6:06

we can write this as $$\frac{-4 i}{(e^{-i x} - e^{i x}) (e^{-i x} + e^{i x})}$$So by letting $u = e^{ix}$ we can write this as $$\frac{-4 i}{(1/u - u) (1/u + u)} = -4i\frac{u^2}{(1 - u^2) (1+u^2)} = \frac{4i}{2 (u^2 + 1)} - \frac{4i}{4 (u + 1)} + \frac{4i}{4 (u - 1)}$$ assuming I did my calculations correctly. Note that this isn't going to combine together in the way you want. Partial fractions are fundamentally property of rational functions of polynomials.