Question

d) $1 + cos x = \frac{sin^2 x}{1 - cos x}$
证：$\frac{sin^2 x}{1 - cos x} = \frac{1 - cos^2 x}{1 - cos x} = 1 + cos x$

Explanation:

Step1: Recall Pythagorean Identity

We know the Pythagorean identity $\sin^{2}x + \cos^{2}x = 1$, which can be rearranged to $\sin^{2}x = 1 - \cos^{2}x$.

Step2: Factor the Right - Hand Side

The expression $1 - \cos^{2}x$ is a difference of squares, so $1 - \cos^{2}x=(1 + \cos x)(1 - \cos x)$.

Step3: Substitute and Simplify

We want to show that $1+\cos x=\frac{\sin^{2}x}{1 - \cos x}$. Substitute $\sin^{2}x = 1 - \cos^{2}x=(1 + \cos x)(1 - \cos x)$ into the right - hand side:
$\frac{\sin^{2}x}{1 - \cos x}=\frac{(1 + \cos x)(1 - \cos x)}{1 - \cos x}$
Since $1-\cos x
eq0$ (if $1 - \cos x = 0$, then $\cos x=1$ and the original equation would be undefined in the form we are manipulating), we can cancel out the common factor of $1 - \cos x$ from the numerator and the denominator.
After canceling, we get $1+\cos x$, which is equal to the left - hand side.

Answer:

The identity $1+\cos x=\frac{\sin^{2}x}{1 - \cos x}$ is proven using the Pythagorean identity $\sin^{2}x=1 - \cos^{2}x$ (and factoring the difference of squares) and simplifying the resulting expression.