$$\operatorname{cosec}\left[2\cot^{-1}(5) + \cos^{-1}\left(\frac{4}{5}\right)\right]$$ is equal to:

We need to find $$\csc\left[2\cot^{-1}(5) + \cos^{-1}\left(\frac{4}{5}\right)\right]$$.

Let $$\theta = \cot^{-1}(5)$$, so $$\tan\theta = \frac{1}{5}$$. Then $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta} = \frac{2/5}{1 - 1/25} = \frac{2/5}{24/25} = \frac{10}{24} = \frac{5}{12}$$.

From $$\tan(2\theta) = \frac{5}{12}$$, we get $$\sin(2\theta) = \frac{5}{13}$$ and $$\cos(2\theta) = \frac{12}{13}$$.

Let $$\phi = \cos^{-1}\left(\frac{4}{5}\right)$$, so $$\cos\phi = \frac{4}{5}$$ and $$\sin\phi = \frac{3}{5}$$.

Now we compute $$\sin(2\theta + \phi) = \sin(2\theta)\cos\phi + \cos(2\theta)\sin\phi = \frac{5}{13} \cdot \frac{4}{5} + \frac{12}{13} \cdot \frac{3}{5} = \frac{4}{13} + \frac{36}{65} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}$$.

Therefore, $$\csc\left[2\cot^{-1}(5) + \cos^{-1}\left(\frac{4}{5}\right)\right] = \frac{1}{\sin(2\theta + \phi)} = \frac{65}{56}$$.