Determine if True (sin(x))/(1+cos(x))=(1-cos(x))/(sin(x))

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

Start on the left side.

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

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

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

Combine.

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

Simplify numerator.

Apply the distributive property.

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

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

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

Reorder factors in $\sin (x) + \sin (x) (- \cos (x))$ .

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

Simplify denominator.

Expand $(1 + \cos (x)) (1 - \cos (x))$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

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

Apply pythagorean identity.

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

Simplify.

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

Multiply by $1$ .

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

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

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

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

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

Cancel the common factors.

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

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

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