Draw a right triangle with angle \(\displaystyle \theta\), "near side" x, and "opposite side" 1. That way \(\displaystyle cot(\theta)=\) "near side"/"opposite side"\(\displaystyle = x/1= x\) so \(\displaystyle sin(cot^{-1} x)= sin(\theta)\)= "opposite side"/"hypotenuse". The "opposite side" is 1 and, by the Pythagorean theorem, the "hypotenuse" is \(\displaystyle \sqrt{1+ x^2}\):

\(\displaystyle sin(cot^{-1}(x))= \frac{1}{\sqrt{1+ x^2}}\).

Now, draw a right triangle with angle \(\displaystyle \phi\), "opposite side" 1 and "near side" \(\displaystyle \sqrt{1+ x^2}\) so that \(\displaystyle tan(\phi)= \frac{1}{\sqrt{1+ x^2}}= sin(cot^{-1}(x))\) and so \(\displaystyle \theta= tan^{-1}(sin(cot^{-1}(x)))\).

\(\displaystyle cos(tan^{-1}(sin(cot^{-1}(x))))\) is the "near side"/"hypotenuse" for this triangle. Since the "near side" is \(\displaystyle \sqrt{1+ x^2}\) and the "opposite side" is 1, use the Pythagorean theorem again to find the hypotenuse of this triangle.