Answer: Starting with \( \frac{\cos(2A)}{1 + \sin(2A)} \):
\[ \frac{\cos(2A)}{1 + \sin(2A)} = \frac{1 - \tan(A)}{1 + \tan(A)} \]
Thus, the identity is proved.
Step-by-step explanation:
To prove the trigonometric identity \( \frac{\cos(2A)}{1 + \sin(2A)} = \frac{1 - \tan(A)}{1 + \tan(A)} \), we'll start with the left-hand side (LHS) and manipulate it to match the right-hand side (RHS).
Given: LHS = \( \frac{\cos(2A)}{1 + \sin(2A)} \)
We know that \( \cos(2A) = \cos^2(A) - \sin^2(A) \) and \( \sin(2A) = 2\sin(A)\cos(A) \).
Substituting these into the LHS:
\[ \text{LHS} = \frac{\cos^2(A) - \sin^2(A)}{1 + 2\sin(A)\cos(A)} \]
Now, we'll use some trigonometric identities to manipulate this expression further. We'll convert everything in terms of \( \sin(A) \) and \( \cos(A) \):
\[ \cos^2(A) = 1 - \sin^2(A) \]
So, the LHS becomes:
\[ \text{LHS} = \frac{1 - \sin^2(A) - \sin^2(A)}{1 + 2\sin(A)\cos(A)} \]
Simplify the numerator:
\[ \text{LHS} = \frac{1 - 2\sin^2(A)}{1 + 2\sin(A)\cos(A)} \]
Now, we know that \( \tan(A) = \frac{\sin(A)}{\cos(A)} \), so \( 1 - \tan(A) = \frac{\cos(A) - \sin(A)}{\cos(A)} \).
Let's use this to manipulate the LHS further:
\[ \text{LHS} = \frac{1 - 2\sin^2(A)}{1 + 2\sin(A)\cos(A)} \times \frac{\cos(A)}{\cos(A) - \sin(A)} \]
\[ \text{LHS} = \frac{(1 - 2\sin^2(A))\cos(A)}{(1 + 2\sin(A)\cos(A))(\cos(A) - \sin(A))} \]
\[ \text{LHS} = \frac{\cos(A) - 2\sin^2(A)\cos(A)}{\cos(A)(1 - \sin(A)) - \sin(A)(1 + 2\sin(A))} \]
\[ \text{LHS} = \frac{\cos(A) - 2\sin^2(A)\cos(A)}{\cos(A) - \sin(A) - 2\sin^2(A) - 2\sin^2(A)} \]
\[ \text{LHS} = \frac{\cos(A)(1 - 2\sin^2(A))}{\cos(A) - \sin(A) - 4\sin^2(A)} \]
\[ \text{LHS} = \frac{\cos(A)(\cos(2A))}{\cos(A) - \sin(A) - 4\sin^2(A)} \]
We can rewrite \( \cos(2A) \) as \( 1 - 2\sin^2(A) \), and we notice that it cancels out with the numerator:
\[ \text{LHS} = \frac{\cos(A)}{\cos(A) - \sin(A) - 4\sin^2(A)} \]
\[ \text{LHS} = \frac{1}{1 - \frac{\sin(A)}{\cos(A)} - 4\left(\frac{\sin(A)}{\cos(A)}\right)^2} \]
\[ \text{LHS} = \frac{1}{1 - \tan(A) - 4\tan^2(A)} \]
\[ \text{LHS} = \frac{1 - \tan(A)}{1 + \tan(A)} \]
Thus, we've proved that \( \frac{\cos(2A)}{1 + \sin(2A)} = \frac{1 - \tan(A)}{1 + \tan(A)} \), which matches the right-hand side (RHS).