Question

prove the identity. (\frac{sin x}{1 - cos x}=csc x+cot x). note that each statement must be based on a rule chosen from the rule menu. to see a detailed description of a rule the right of the rule.

Explanation:

Step1: Multiply by conjugate

Multiply the left - hand side $\frac{\sin x}{1 - \cos x}$ by $\frac{1+\cos x}{1 + \cos x}$.
$\frac{\sin x}{1 - \cos x}\times\frac{1+\cos x}{1+\cos x}=\frac{\sin x(1 + \cos x)}{1-\cos^{2}x}$

Step2: Use Pythagorean identity

Recall the Pythagorean identity $\sin^{2}x+\cos^{2}x = 1$, so $1-\cos^{2}x=\sin^{2}x$. Then we have $\frac{\sin x(1 + \cos x)}{\sin^{2}x}$.

Step3: Simplify the fraction

$\frac{\sin x(1 + \cos x)}{\sin^{2}x}=\frac{1+\cos x}{\sin x}$.

Step4: Split the fraction

$\frac{1+\cos x}{\sin x}=\frac{1}{\sin x}+\frac{\cos x}{\sin x}$.

Step5: Use trigonometric identities

Since $\frac{1}{\sin x}=\csc x$ and $\frac{\cos x}{\sin x}=\cot x$, we get $\csc x+\cot x$.

Answer:

The identity $\frac{\sin x}{1 - \cos x}=\csc x+\cot x$ is proven.