Trigonometry Equation

Problem 119: Verify the identity:

(1) $\begin{equation*} \frac{\cos^2 x}{1-\sin x} - \frac{\cos^2 x}{1+\sin x} = 2 \sin x. \end{equation*}$

Solution: If we want to verify this equation, we need to pick a side in order to prove the other side. I will start with the left-hand side $(L)$ and show that it is equal to the right-hand side $(R)$ . That is,

(2) $\begin{align*} L = \frac{\cos^2 x}{1-\sin x} - \frac{\cos^2 x}{1+\sin x} & = \frac{\cos^2 x (1+\sin x) - \cos^2 x (1-\sin x)}{(1-\sin x)(1+\sin x)} \\ & = \frac{ \bcancel{\cos^2 x} + \cos^2x \sin x - \bcancel{\cos^2x} + \cos^2 x \sin x}{(1-\sin x)(1+\sin x)} \\& = \frac{2\cos^2 x \sin x}{\underbrace{1-\sin^2 x}_{= \ \cos^2 x}} \\ & = \frac{2\cos^2 x \sin x}{\cos^2 x} \\ & = 2 \sin x = R. \end{align*}$

Hence, $L = R$ as desired.

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