1. Given function

   $y = \sec(\tan^{-1} x)$

   We need to find $\dfrac{dy}{dx}$ at $x=1$.

2. Simplify the function

   Let $\theta = \tan^{-1}x$ Then $\tan\theta = x$

   Using the identity $\sec^2\theta = 1 + \tan^2\theta$ we get $\sec\theta = \sqrt{1+\tan^2\theta} = \sqrt{1+x^2}$

   Hence, $y = \sqrt{1+x^2}$

3. Differentiate

   $y = (1+x^2)^{1/2}$

   Therefore, $\frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-1/2}(2x) = \frac{x}{\sqrt{1+x^2}}$

4. Evaluate at $x=1$

   $\left.\frac{dy}{dx}\right|_{x=1} = \frac{1}{\sqrt{1+1}} = \frac{1}{\sqrt{2}}$

5. Option check

   $\frac{1}{\sqrt{2}}$ matches Option A.

Therefore, the required value is $\boxed{\frac{1}{\sqrt{2}}}$