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

Verify the identity by following steps: - step0: Verify: \(\frac{1+\sin\left(y\right)}{\cos\left(y\right)}=\frac{2}{\cos\left(y\right)}\) - step1: Choose a side to work on: \(\frac{1}{\cos\left(y\right)}+\frac{\sin\left(y\right)}{\cos\left(y\right)}=\frac{2}{\cos\left(y\right)}\) - step2: Verify the identity: \(\textrm{false}\) Determine whether the expression \( \frac{\cos y}{1+\sin y}=\frac{1+\sin y}{\cos y} \) is always true. Verify the identity by following steps: - step0: Verify: \(\frac{\cos\left(y\right)}{1+\sin\left(y\right)}=\frac{1+\sin\left(y\right)}{\cos\left(y\right)}\) - step1: Choose the other side to work on: \(\frac{\cos\left(y\right)}{1+\sin\left(y\right)}=\frac{1}{\cos\left(y\right)}+\frac{\sin\left(y\right)}{\cos\left(y\right)}\) - step2: Verify the identity: \(\textrm{false}\) The statement \( \frac{\cos y}{1+\sin y}=\frac{1+\sin y}{\cos y}=\frac{2}{\cos y} \) is not true. 1. **First part:** \( \frac{\cos y}{1+\sin y} = \frac{1+\sin y}{\cos y} \) This equality is false. To see why, we can cross-multiply: \[ \cos^2 y = (1 + \sin y)(1 + \sin y) = 1 + 2\sin y + \sin^2 y \] This simplifies to: \[ \cos^2 y = 1 + 2\sin y + \sin^2 y \] Using the Pythagorean identity \( \cos^2 y + \sin^2 y = 1 \), we can substitute \( \cos^2 y \) with \( 1 - \sin^2 y \): \[ 1 - \sin^2 y = 1 + 2\sin y + \sin^2 y \] Rearranging gives: \[ -\sin^2 y - 2\sin y = 0 \] Factoring out \(-\sin y\): \[ -\sin y(\sin y + 2) = 0 \] This implies \( \sin y = 0 \) or \( \sin y = -2 \) (which is not possible). Thus, the equality does not hold for all \( y \). 2. **Second part:** \( \frac{1+\sin y}{\cos y} = \frac{2}{\cos y} \) This equality is also false. Cross-multiplying gives: \[ 1 + \sin y = 2 \] This simplifies to: \[ \sin y = 1 \] This is only true for specific values of \( y \) (e.g., \( y = \frac{\pi}{2} + 2k\pi \) for integers \( k \)), not for all \( y \). In conclusion, the original statement is not universally true.