MHB Home work help: proving a trigonometric identity

AI Thread Summary
The discussion centers on proving the trigonometric identity 1/(1 + cos(θ)) = csc²(θ) - csc(θ)cot(θ). A participant emphasizes the importance of converting all terms to sine and cosine for simplification. The proof involves manipulating the right-hand side to show it equals the left-hand side. The initial step of rewriting the equation is highlighted as crucial for solving the problem. Overall, the exchange illustrates effective methods for tackling trigonometric identities.
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___________ =csc2\theta-csc\thetacot\theta
1+cos\theta
 
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Re: home work help

Are you asking how to prove the identity \frac{1}{1+ cos(\theta)}= csc^2(\theta)- csc(\theta)cot(\theta)?
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:
\frac{1}{1+ cos(\theta)}= \frac{1}{sin^2(\theta)}- \frac{1}{sin(\theta)}\frac{cos(\theta)}{sin(\theta)}
 
816318 said:
\frac{1}{1 + \cos\theta} \;=\; \csc^2\theta-\csc\theta\cot\theta []/size]

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}

. . . . . =\;\;\frac{1-\cos\theta}{(1-\cos\theta)(1+\cos\theta)} \;\;=\;\;\frac{1}{1+\cos\theta} \;\;=\;\; LHS
 
$$\frac{1}{1+\cos(\theta)}\cdot\frac{1-\cos(\theta)}{1-\cos(\theta)}=\frac{1-\cos(\theta)}{\sin^2(\theta)}=\csc^2(\theta)-\csc(\theta)\cot(\theta)$$
 
Re: home work help

HallsofIvy said:
Are you asking how to prove the identity \frac{1}{1+ cos(\theta)}= csc^2(\theta)- csc(\theta)cot(\theta)?
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:
\frac{1}{1+ cos(\theta)}= \frac{1}{sin^2(\theta)}- \frac{1}{sin(\theta)}\frac{cos(\theta)}{sin(\theta)}

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