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.

  • 4
    $\begingroup$ 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 $A = \cos(x), B = \sin(x)$. $\endgroup$
    – xbh
    Sep 12, 2018 at 6:06

2 Answers 2


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.

  • $\begingroup$ +1 I didn't notice this way for a while! $\endgroup$
    – Mikasa
    Sep 12, 2018 at 6:08
  • $\begingroup$ @Jean-ClaudeArbaut a proof of impossibility is given by the OP in his post, so I feel no need to repeat it. There is no answer to the problem if you define an answer to be a PFD of the type the OP wants. I figured it would thus be instructive to present the OP with as close to a PFD as I think you can get. Definitely applicable - a decomposition like this is standard in complex analysis, e.g. in contour integration. $\endgroup$ Sep 12, 2018 at 6:35

\begin{align} \sin \sin x + \cos x \cos x &= 1 \\ \dfrac{\sin x \sin x}{\sin x \cos x} + \dfrac{\cos x \cos x}{\sin x \cos x} &= \dfrac{1}{\sin x \cos x} \\ \dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} &= \dfrac{1}{\sin x \cos x} \\ \end{align}


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