Step 1: Simplify \( \cot(\cos^{-1} x) \)
Let \( \theta = \cos^{-1} x \), so that \( \cos \theta = x \). Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \sin \theta \):
\[
\sin^2 \theta = 1 - x^2 \quad \Rightarrow \quad \sin \theta = \sqrt{1 - x^2}.
\]
Now, we can compute \( \cot(\cos^{-1} x) \) as follows:
\[
\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{\sqrt{1 - x^2}}.
\]
Thus:
\[
\cot(\cos^{-1} x) = \frac{x}{\sqrt{1 - x^2}}.
\]
Step 2: Simplify \( \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right) \)
Let \( \phi = \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \). Then \( \tan \phi = \frac{a}{\sqrt{b^2 - a^2}} \). Using the identity \( \sec^2 \phi = 1 + \tan^2 \phi \), we can find \( \sec \phi \):
\[
\sec^2 \phi = 1 + \left( \frac{a}{\sqrt{b^2 - a^2}} \right)^2 = 1 + \frac{a^2}{b^2 - a^2}.
\]
Simplifying:
\[
\sec^2 \phi = \frac{b^2 - a^2 + a^2}{b^2 - a^2} = \frac{b^2}{b^2 - a^2}.
\]
Thus:
\[
\sec \phi = \frac{b}{\sqrt{b^2 - a^2}}.
\]
Step 3: Equate the two sides
We are given the equation:
\[
\cot(\cos^{-1} x) = \sec \left( \tan^{-1} \left( \frac{a}{\sqrt{b^2 - a^2}} \right) \right).
\]
From Step 1, we have:
\[
\frac{x}{\sqrt{1 - x^2}} = \frac{b}{\sqrt{b^2 - a^2}}.
\]
Now, solving for \( x \), we cross-multiply:
\[
x \sqrt{b^2 - a^2} = b \sqrt{1 - x^2}.
\]
Squaring both sides:
\[
x^2 (b^2 - a^2) = b^2 (1 - x^2).
\]
Expanding both sides:
\[
x^2 b^2 - x^2 a^2 = b^2 - b^2 x^2.
\]
Rearranging terms:
\[
x^2 (b^2 - a^2 + b^2) = b^2.
\]
Thus:
\[
x^2 (2b^2 - a^2) = b^2.
\]
Solving for \( x^2 \):
\[
x^2 = \frac{b^2}{2b^2 - a^2}.
\]
Finally, taking the square root:
\[
x = \frac{b}{\sqrt{2b^2 - a^2}}.
\]
Final Answer:
The value of \( x \) is:
\( \frac{b}{\sqrt{2b^2 - a^2}} \)