Home work help: proving a trigonometric identity

[816318]

1
___________ =csc2\theta-csc\thetacot\theta
1+cos\theta

 

[HallsofIvy]

Re: home work help

Are you asking how to prove the identity [tex]\frac{1}{1+ cos(\theta)}= csc^2(\theta)- csc(\theta)cot(\theta)[/tex]?
Also you labeled this "home work" help. Did you make any attempt at this yourself or do you just want someone to do your home work for you?

My first reaction for a problem like this is always to change every thing to sine and cosine:
[tex]\frac{1}{1+ cos(\theta)}= \frac{1}{sin^2(\theta)}- \frac{1}{sin(\theta)}\frac{cos(\theta)}{sin(\theta)}[/tex]

 

[soroban]

[tex]\frac{1}{1 + \cos\theta} \;=\; \csc^2\theta-\csc\theta\cot\theta[/tex] []/size]

[tex]RHS \;\;=\;\;\frac{1}{\sin^2\theta} - \frac{1}{\sin\theta}\frac{\cos\theta}{\sin\theta} \;\;=\;\;\frac{1-\cos\theta}{\sin^2\theta} \;\;=\;\;\frac{1-\cos\theta}{1-\cos^2\theta} [/tex]

. . . . . [tex]=\;\;\frac{1-\cos\theta}{(1-\cos\theta)(1+\cos\theta)} \;\;=\;\;\frac{1}{1+\cos\theta} \;\;=\;\; LHS[/tex]

 

[Greg]

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

 

[816318]

Re: home work help

Are you asking how to prove the identity [tex]\frac{1}{1+ cos(\theta)}= csc^2(\theta)- csc(\theta)cot(\theta)[/tex]?
Also you labeled this "home work" help. Did you make any attempt at this yourself or do you just want someone to do your home work for you?

My first reaction for a problem like this is always to change every thing to sine and cosine:
[tex]\frac{1}{1+ cos(\theta)}= \frac{1}{sin^2(\theta)}- \frac{1}{sin(\theta)}\frac{cos(\theta)}{sin(\theta)}[/tex]

Thanks buddy, the first step was all that was needed to solve the rest!