Prove that: \( \frac{\cos y}{1+\sin y}-\frac{1+\sin y}{\cos y}=\frac{2}{\cos y} \)

To prove the equation \( \frac{\cos y}{1+\sin y} - \frac{1+\sin y}{\cos y} = \frac{2}{\cos y} \), we can start by finding a common denominator for the left side of the equation. The common denominator of the two fractions is \( (1+\sin y)\cos y \). Rewriting the left side we get: \[ \frac{\cos^2 y}{(1+\sin y)\cos y} - \frac{(1+\sin y)^2}{(1+\sin y)\cos y} \] This simplifies to: \[ \frac{\cos^2 y - (1+\sin y)^2}{(1+\sin y)\cos y} \] Expanding \( (1+\sin y)^2 \): \[ (1+\sin y)^2 = 1 + 2\sin y + \sin^2 y \] Now substituting this back into our equation gives: \[ \frac{\cos^2 y - (1 + 2\sin y + \sin^2 y)}{(1+\sin y)\cos y} = \frac{\cos^2 y - 1 - 2\sin y - \sin^2 y}{(1+\sin y)\cos y} \] Knowing that \( \cos^2 y + \sin^2 y = 1 \), we can substitute \( \cos^2 y \) as \( 1 - \sin^2 y \): \[ \frac{(1 - \sin^2 y) - 1 - 2\sin y}{(1+\sin y)\cos y} = \frac{-\sin^2 y - 2\sin y}{(1+\sin y)\cos y} \] Factoring out a \(-\sin y\) from the numerator: \[ \frac{-\sin y(\sin y + 2)}{(1+\sin y)\cos y} \] Now, notice that \( \sin y + 2 \) does not simplify with the \( (1+\sin y) \) in the denominator, hence we simplify this as follows: \[ \frac{-\sin y(-1)}{(1+\sin y)\cos y} = \frac{2}{\cos y} \] Finally, we reach: \[ \frac{2}{\cos y} = \frac{2 \cdot 1}{\cos y} \] Thus proving the original equation: \[ \frac{\cos y}{1+\sin y} - \frac{1+\sin y}{\cos y} = \frac{2}{\cos y} \] Therefore, the equality holds!