If $y = \cot^{-1}\left(\dfrac{1 + \sin 5x}{\cos 5x}\right)$, then the value of $\dfrac{dy}{dx}$ is

Given $y = \cot^{-1}\left(\dfrac{1 + \sin 5x}{\cos 5x}\right)$

We can simplify the expression inside the inverse trigonometric function using half-angle formulas:

$\dfrac{1 + \sin 5x}{\cos 5x} = \dfrac{\left(\cos \dfrac{5x}{2} + \sin \dfrac{5x}{2}\right)^2}{\cos^2 \dfrac{5x}{2} - \sin^2 \dfrac{5x}{2}}$

$\Rightarrow \dfrac{1 + \sin 5x}{\cos 5x} = \dfrac{\cos \dfrac{5x}{2} + \sin \dfrac{5x}{2}}{\cos \dfrac{5x}{2} - \sin \dfrac{5x}{2}}$

Dividing the numerator and the denominator by $\cos \dfrac{5x}{2}$:

$\dfrac{1 + \sin 5x}{\cos 5x} = \dfrac{1 + \tan \dfrac{5x}{2}}{1 - \tan \dfrac{5x}{2}} = \tan\left(\dfrac{\pi}{4} + \dfrac{5x}{2}\right)$

Using the complementary angle identity $\tan \theta = \cot\left(\dfrac{\pi}{2} - \theta\right)$:

$\tan\left(\dfrac{\pi}{4} + \dfrac{5x}{2}\right) = \cot\left(\dfrac{\pi}{2} - \left(\dfrac{\pi}{4} + \dfrac{5x}{2}\right)\right) = \cot\left(\dfrac{\pi}{4} - \dfrac{5x}{2}\right)$

Substituting this back into the original equation:

$y = \cot^{-1}\left[\cot\left(\dfrac{\pi}{4} - \dfrac{5x}{2}\right)\right]$

$y = \dfrac{\pi}{4} - \dfrac{5x}{2}$

Differentiating with respect to $x$:

$\dfrac{dy}{dx} = \dfrac{d}{dx}\left(\dfrac{\pi}{4} - \dfrac{5x}{2}\right) = -\dfrac{5}{2}$

Answer: $-\dfrac{5}{2}$