# How do you simplify cotB + (sinB / (1+cosB))?

##### 2 Answers
Nov 5, 2015

$\csc \beta$

#### Explanation:

$\cot \beta$+$\sin \frac{\beta}{1 + \cos \beta}$

1) Turn everything into sin and cos.

$\cos \frac{\beta}{\sin} \beta + \sin \frac{\beta}{1 + \cos \beta}$

2) Use LCD and create a common numerator.

$\frac{1 + \cos \beta}{1 + \cos \beta} \cdot \cos \frac{\beta}{\sin} \beta + \sin \frac{\beta}{1 + \cos \beta} \cdot \sin \frac{\beta}{\sin} \beta$

$\frac{\left(1 + \cos \beta\right) \cos \beta + {\sin}^{2} \beta}{\sin \beta \left(1 + \cos \beta\right)}$

3) Distibute $\cos \beta$ in the denominator.

$\frac{\cos \beta + \left({\cos}^{2} \beta + {\sin}^{2} \beta\right)}{\sin \beta \left(1 + \cos \beta\right)}$

4) Pythagorean Identity in the denominator. Then cancel out matching denominator and numerator.

$\frac{\cancel{\cos \beta + \left(1\right)}}{\sin \beta \left(\cancel{1 + \cos \beta}\right)}$

5) Left with $\frac{1}{\sin} \beta$ = $\csc \beta$

Nov 5, 2015

$\csc \left(B\right)$

#### Explanation:

$\cot \left(B\right) + \frac{\sin \left(B\right)}{1 + \cos \left(B\right)} = \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{\sin \left(B\right)}{1 + \cos \left(B\right)} \frac{1 - \cos \left(B\right)}{1 - \cos \left(B\right)}$

$= \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{\sin \left(B\right) \left[1 - \cos \left(B\right)\right]}{1 - {\cos}^{2} \left(B\right)}$

$= \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{\sin \left(B\right) \left[1 - \cos \left(B\right)\right]}{{\sin}^{2} \left(B\right)}$

$= \frac{\cos \left(B\right)}{\sin \left(B\right)} + \frac{1 - \cos \left(B\right)}{\sin \left(B\right)}$

$= \frac{1}{\sin \left(B\right)}$

$= \csc \left(B\right)$