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

Here's an approach which is relatively light on computation.
As in the comments, we observe that $A^2 = -(x^2 + y^2 + z^2)I$.  It follows that any rational function of $A$ can be written in the form $p I + q A$ for some $p,q$.  In other words, there necessarily exist numbers $p,q$ such that
$$
(M-I)^{-1} = ([\cos(a) - 1]I + \sin(a)A)^{-1} = p I + q A
$$
Now, expand the product $(M-I)(pI + qA),$ and solve for the $p$ and $q$ that cause this product to equal $I$.  In particular, we have
$$
(M - I)(pI + qA) = ([\cos(a) - 1]I + \sin(a)A)(pI + qA)\\
= p(\cos(a) - 1)I + (q[\cos(a) - 1] + p\sin(a))A + \sin(a)qA^2\\
= \left[p(\cos(a) - 1) - \sin(a)(x^2 + y^2 + z^2)q \right]I +  \left[q(\cos(a) - 1) + p\sin(a)\right]A.
$$
Thus, it suffices to find $p,q$ that solve the system of equations
$$
\begin{array}{ccccccc}
(\cos(a) - 1) &p &- &\sin(a)(x^2 + y^2 + z^2)&q &= &1\\
\sin(a) &p &+ &(\cos(a) - 1)&q &= &0.
\end{array}
$$
Solving this system yields
$$
p = \frac{1 - \cos(a)}{D}, \quad q = \frac{\sin(a)}{D}
$$
where $D$ is the determinant of the coefficient matrix of this system, that is
$$
D = \sin^2(a)(x^2 + y^2 + z^2) + (\cos(a) - 1)^2.
$$
I see no general way of simplifying the resulting expression.
