Verify the Identity (1-cos(x))/(sin(x))+(sin(x))/(1-cos(x))=2csc(x)

$\frac{1 - \cos (x)}{\sin (x)} + \frac{\sin (x)}{1 - \cos (x)} = 2 \csc (x)$

Start on the left side.

$\frac{1 - \cos (x)}{\sin (x)} + \frac{\sin (x)}{1 - \cos (x)}$

Add fractions.

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To write $\frac{1 - \cos (x)}{\sin (x)}$ as a fraction with a common denominator, multiply by $\frac{1 - \cos (x)}{1 - \cos (x)}$ .

$\frac{1 - \cos (x)}{\sin (x)} \cdot \frac{1 - \cos (x)}{1 - \cos (x)} + \frac{\sin (x)}{1 - \cos (x)}$

To write $\frac{\sin (x)}{1 - \cos (x)}$ as a fraction with a common denominator, multiply by $\frac{\sin (x)}{\sin (x)}$ .

$\frac{1 - \cos (x)}{\sin (x)} \cdot \frac{1 - \cos (x)}{1 - \cos (x)} + \frac{\sin (x)}{1 - \cos (x)} \cdot \frac{\sin (x)}{\sin (x)}$

Write each expression with a common denominator of $\sin (x) (1 - \cos (x))$ , by multiplying each by an appropriate factor of $1$ .

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Combine.

$\frac{(1 - \cos (x)) (1 - \cos (x))}{\sin (x) (1 - \cos (x))} + \frac{\sin (x)}{1 - \cos (x)} \cdot \frac{\sin (x)}{\sin (x)}$

Combine.

$\frac{(1 - \cos (x)) (1 - \cos (x))}{\sin (x) (1 - \cos (x))} + \frac{\sin (x) \sin (x)}{(1 - \cos (x)) \sin (x)}$

Reorder the factors of $(1 - \cos (x)) \sin (x)$ .

$\frac{(1 - \cos (x)) (1 - \cos (x))}{\sin (x) (1 - \cos (x))} + \frac{\sin (x) \sin (x)}{\sin (x) (1 - \cos (x))}$

Combine the numerators over the common denominator.

$\frac{(1 - \cos (x)) (1 - \cos (x)) + \sin (x) \sin (x)}{\sin (x) (1 - \cos (x))}$

Simplify each term.

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

Simplify denominator.

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Apply the distributive property.

$\frac{1 - 2 \cos (x) + \cos^{2} (x) + \sin^{2} (x)}{\sin (x) \cdot 1 + \sin (x) (- \cos (x))}$

Multiply $\sin (x)$ by $1$ .

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

Simplify each term.

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

Apply pythagorean identity.

$\frac{1 - 2 \cos (x) + 1}{\sin (x) - \sin (x) \cos (x)}$

Simplify.

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Simplify the numerator.

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Add $1$ and $1$ .

$\frac{2 - 2 \cos (x)}{\sin (x) - \sin (x) \cos (x)}$

Factor $2$ out of $2 - 2 \cos (x)$ .

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Factor $2$ out of $2$ .

$\frac{2 (1) - 2 \cos (x)}{\sin (x) - \sin (x) \cos (x)}$

Factor $2$ out of $- 2 \cos (x)$ .

$\frac{2 (1) + 2 (- \cos (x))}{\sin (x) - \sin (x) \cos (x)}$

Factor $2$ out of $2 (1) + 2 (- \cos (x))$ .

$\frac{2 (1 - \cos (x))}{\sin (x) - \sin (x) \cos (x)}$

Factor $\sin (x)$ out of $\sin (x) - \sin (x) \cos (x)$ .

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Multiply by $1$ .

$\frac{2 (1 - \cos (x))}{\sin (x) \cdot 1 - \sin (x) \cos (x)}$

Factor $\sin (x)$ out of $- \sin (x) \cos (x)$ .

$\frac{2 (1 - \cos (x))}{\sin (x) \cdot 1 + \sin (x) (- \cos (x))}$

Factor $\sin (x)$ out of $\sin (x) \cdot 1 + \sin (x) (- \cos (x))$ .

$\frac{2 (1 - \cos (x))}{\sin (x) (1 - \cos (x))}$

Cancel the common factor of $1 - \cos (x)$ .

$\frac{2}{\sin (x)}$

Now consider the right side of the equation.

$2 \csc (x)$

Apply the reciprocal identity to $\csc (x)$ .

$2 \frac{1}{\sin (x)}$

Combine $2$ and $\frac{1}{\sin (x)}$ .

$\frac{2}{\sin (x)}$

Because the two sides have been shown to be equivalent, the equation is an identity.

$\frac{1 - \cos (x)}{\sin (x)} + \frac{\sin (x)}{1 - \cos (x)} = 2 \csc (x)$ is an identity

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