Inverse of a skew-symmetric matrix For $a, x, y, z \in \mathbb R$, let $$M= \left(
  \begin{array}{cccc}
    \cos(a)  & \sin(a) \, x & \sin(a)\,  y & \sin(a) \, z \\
    -\sin(a) \, x & \cos(a) & \sin(a) \,z & -\sin(a)\, y \\
   -\sin(a) \, y & -\sin(a) \, z & \cos(a) & \sin(a) \, x \\
  -\sin(a) \, z & \sin(a) \, y & -\sin(a)\, x & \cos(a)
  \end{array}
\right).$$ 
Without doing much calculation, why the matrix $M-I_4$ is invertible and why its inverse is given $$(M-I_4)^{-1}=\,\frac{-1}{2} I_{4} - \frac{\cot(\sqrt{x^2+y^2+z^2} )}{2\sqrt{x^2+y^2+z^2}} A,$$
where $A$ is the skew-symmetric matrix given by 
$$A=\left(
  \begin{array}{cccc}
    0 & x & y &  z \\
    -x & 0 & z & -y \\
    -y & -z & 0 & x \\
    -z & y & -x & 0
  \end{array}
\right).$$
Thank you in advance

$A$ has the form $A = \begin{pmatrix}J&L\\-L&J\end{pmatrix}$, where $JL+LJ = 0$, $J^2 = -x^2I$, and $L^2 = (y^2+z^2)I$. Therefore,
$$
A^2 = \begin{pmatrix}J&L\\-L&J\end{pmatrix}\begin{pmatrix}J&L\\-L&J\end{pmatrix} = \begin{pmatrix}J^2-L^2&0\\0&J^2-L^2\end{pmatrix} = -r^2I_4,
$$
where $r := \sqrt{x^2+y^2+z^2}$. Hence, for $z\in\mathbb C$ we have $(A-z)(A+z) = A^2-z^2I = -(r^2+z^2)I$ and thus
$$
(A-z)^{-1} = -\frac{A+z}{r^2+z^2}.
$$
Now, as I wrote in the comments, we have $M-I = (\cos(a)-1)I + \sin(a)A$. So, for $a = 2k\pi$ the matrix $M-I$ is singular, whereas for $a=(2k+1)\pi$ we have $M-I = -2I$. If $a$ is not one of these values,
$$
(M-I)^{-1} = \left[\sin(a)\left(A - \frac{1-\cos(a)}{\sin(a)}\right)\right]^{-1} = \frac 1{\sin(a)}\cdot\frac{A+\alpha}{r^2+\alpha^2},
$$
where $\alpha = \frac{1-\cos(a)}{\sin(a)}$.
