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

We start with the left‐hand side (LHS) of the equation: \[ \frac{\cos y}{1+\sin y}-\frac{1+\sin y}{\cos y}. \] Our goal is to simplify this expression. ### Step 1. Write the Expression with a Common Denominator The denominators are \(1+\sin y\) and \(\cos y\). A common denominator is \(\cos y (1+\sin y)\). Thus, we write: \[ \frac{\cos y}{1+\sin y}-\frac{1+\sin y}{\cos y} = \frac{\cos^2 y}{\cos y(1+\sin y)} - \frac{(1+\sin y)^2}{\cos y(1+\sin y)}. \] ### Step 2. Combine the Fractions Combine the two terms over the common denominator: \[ \frac{\cos^2 y - (1+\sin y)^2}{\cos y(1+\sin y)}. \] ### Step 3. Expand and Simplify the Numerator First, expand the square in the numerator: \[ (1+\sin y)^2 = 1 + 2\sin y + \sin^2 y. \] Now the numerator becomes: \[ \cos^2 y - (1 + 2\sin y + \sin^2 y) = \cos^2 y - 1 - 2\sin y - \sin^2 y. \] Recall the Pythagorean identity: \[ \cos^2 y = 1 - \sin^2 y. \] Substitute this into the numerator: \[ 1 - \sin^2 y - 1 - 2\sin y - \sin^2 y = -2\sin y - 2\sin^2 y. \] Factor out the common factor \(-2\sin y\): \[ -2\sin y(1+\sin y). \] ### Step 4. Cancel the Common Factor The entire expression is now: \[ \frac{-2\sin y(1+\sin y)}{\cos y(1+\sin y)}. \] Assuming \(1+\sin y\neq 0\) (i.e. \(\sin y\neq -1\)), cancel the common factor \((1+\sin y)\): \[ -\frac{2\sin y}{\cos y}. \] ### Step 5. Compare with the Right-hand Side The original equation to prove is: \[ \frac{\cos y}{1+\sin y}-\frac{1+\sin y}{\cos y} = \frac{2}{\cos y}. \] However, we have shown: \[ \frac{\cos y}{1+\sin y}-\frac{1+\sin y}{\cos y} = -\frac{2\sin y}{\cos y}. \] For these two expressions to be equal, we would require: \[ -\frac{2\sin y}{\cos y} = \frac{2}{\cos y}. \] Multiplying both sides by \(\cos y\) (assuming \(\cos y\neq 0\)) leads to: \[ -2\sin y = 2. \] Thus, \[ \sin y = -1. \] ### Conclusion The simplified form of the left‐hand side is: \[ -\frac{2\sin y}{\cos y}. \] This is equal to \(\frac{2}{\cos y}\) only when \(\sin y = -1\). In general, the identity \[ \frac{\cos y}{1+\sin y}-\frac{1+\sin y}{\cos y}=\frac{2}{\cos y} \] does not hold for all values of \(y\).