Question

Prove the following identity. Make sure to include all steps taken. 20%7D%7Bsin%5Ctheta%7D%2B%5Cfrac%7Bsin%5Ctheta%7D%7B1%2Bcos%5Ctheta%7D%3D2csc%5Ctheta" id="TexFormula1" title="\frac{1+cos\theta }{sin\theta}+\frac{sin\theta}{1+cos\theta}=2csc\theta" alt="\frac{1+cos\theta }{sin\theta}+\frac{sin\theta}{1+cos\theta}=2csc\theta" align="absmiddle" class="latex-formula">

Answer 1

Answer:

See below.

Step-by-step explanation:

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

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

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

$\frac{(1+2cos(\theta)+cos^2(\theta)+sin^2(\theta))}{sin(\theta)(1+cos(\theta))} =2csc(\theta)$

Recall the identities:

$sin^2(\theta)+cos^2(\theta)=1$

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

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

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

$\frac{2}{sin(\theta)} =2csc(\theta)$

$2csc(\theta)=2csc(\theta)$