If $\cot \left(\operatorname{Cos}^{-1} \mathrm{x}\right)=\sec \left\{\operatorname{Tan}^{-1}\left(\frac{\mathrm{a}}{\sqrt{\mathrm{b}^2-\mathrm{a}^2}}\right)\right\}, \mathrm{b}>\mathrm{a}$, then $\mathrm{x}=$

• Correct answer
  A

  $\frac{\mathrm{b}}{\sqrt{2 \mathrm{~b}^2-\mathrm{a}^2}}$
• B

  $\frac{a}{\sqrt{2 b^2-a^2}}$
• C

  $\frac{\sqrt{b^2-a^2}}{a}$
• D

  $\frac{\sqrt{b^2-a^2}}{b}$

Hence, the correct answer is option: 'A'.