Step-by-step explanation:

To prove the given trigonometric identity, we can start with the left-hand side (LHS) and manipulate it step by step to obtain the right-hand side (RHS).

Starting with the LHS:

\[1 + \frac{\cos^2(A)}{1 - \sin(A)}\]

To simplify, we can use the identity \(\sin^2(A) + \cos^2(A) = 1\):

\[= \frac{1 - \sin^2(A) + \cos^2(A)}{1 - \sin(A)}\]

Combine like terms:

\[= \frac{1 - \sin^2(A)}{1 - \sin(A)}\]

Factor out \(-\sin(A)\) from the numerator:

\[= \frac{-\sin(A)(\sin(A) - 1)}{1 - \sin(A)}\]

Cancel common factor \(-(sin(A) - 1)\):

\[= -\sin(A)\]

Now, let's simplify the RHS:

\[\tan^4(A) + \sec^4(A)\]

Using the identity \(\sec^2(A) = 1 + \tan^2(A)\), we can express \(\sec^4(A)\) in terms of \(\tan(A)\):

\[\sec^4(A) = (1 + \tan^2(A))^2\]

Expand and simplify:

\[= 1 + 2\tan^2(A) + \tan^4(A)\]

Now, add \(\tan^4(A)\) and \(\sec^4(A)\):

\[= \tan^4(A) + 1 + 2\tan^2(A) + \tan^4(A)\]

Combine like terms:

\[= 2\tan^4(A) + 2\tan^2(A) + 1\]

Now, we have:

\[LHS = -\sin(A)\]

\[RHS = 2\tan^4(A) + 2\tan^2(A) + 1\]

The LHS is not equal to the RHS, and it seems there might be a mistake in the given identity or in the steps of the proof. Please double-check the identity or provide additional information for clarification.