# Thread: Prove given trig identity

1. ## Prove given trig identity

I desperately need help on how to prove this...if someone can help me with the steps pleaseeeee....I believe you need to go from LHS to RHS...

sinθ/(1-cosθ )-(1+cosθ)/sinθ =0

2. This one is one of those times where you just need to start "doing stuff". If you are able to go from the RHS to the LHS, well heck I would be darn sure impressed:

$\displaystyle \frac{\sin \Theta }{1-\cos \Theta }-\frac{1+\cos \Theta }{\sin \Theta }=0\Rightarrow$

$\displaystyle \frac{\sin ^{2}\Theta -(1-\cos^{2} \Theta )}{(1-\cos \Theta )(\sin \Theta )}=0\Rightarrow$

. . .See if you can't take it from there.

3. LHS = sinθ/(1 - cosθ) - (1 + cosθ)/sinθ
. . . .= sin²θ/[sinθ(1 - cosθ)] - (1 - cos²θ)/[sinθ(1 - cosθ)] (get LCD)
. . . .= (sin²θ + cos²θ - 1)/[sinθ(1 - cosθ)]
. . . .= (1 - 1)/[sinθ(1 - cosθ)] (sin²θ + cos²θ = 1)
. . . .= 0/[sinθ(1 - cosθ)]
. . . .= 0
. . . .= RHS.

Is this correct? If it is, I guess I'm looking for more of an explanation of what I did....I'm thinking it's right, just confused...Thanks

4. You should have posted this in the Trigonometry forum.

$\displaystyle \dfrac{\sin{\theta}}{1-\cos\theta}-\dfrac{1+\cos\theta}{\sin\theta}$ $\displaystyle = \dfrac{\sin^2{\theta}-\left(1+\cos\theta\right)\left(1-\cos\theta\right)}{\left(\sin\theta\right)\left(1-\cos\theta\right)} = \dfrac{\sin^2{\theta}-\left(1-\cos^2\theta\right)}{\left(\sin\theta\right)\left( 1-\cos\theta\right)} = \dfrac{\sin^2\theta+\cos^2\theta-1}{\left(\sin\theta\right)\left(1-\cos\theta\right)}$, then recall Pythagoras & you are done.

Is this correct?
Correct!

6. I'm not sure why everyone's giving such complicated proofs. The original equation can be simplified thus

$\displaystyle \frac{\sin^2\left(\theta\right)}{1-\cos\left(\theta\right)}-\left(1+\cos\left(\theta\right)\right)=0$
$\displaystyle \Rightarrow \sin^2\left(\theta\right)=\left(1+\cos\left(\theta \right)\right)\left(1-\cos\left(\theta\right)\right)$
$\displaystyle \Rightarrow \sin^2\left(\theta\right)=1-\cos^2\left(\theta\right)$
$\displaystyle \Rightarrow \sin^2\left(\theta\right)+\cos^2\left(\theta\right )=1$

which is the Pythagorean identity.

7. Originally Posted by TheCoffeeMachine
Correct!
Is there an explanation of what i did...Like if I were to explain this problem.....any help for a step by step explanation?

8. Originally Posted by lovek323
I'm not sure why everyone's giving such complicated proofs. The original equation can be simplified thus

$\displaystyle \frac{\sin^2\left(\theta\right)}{1-\cos\left(\theta\right)}-\left(1+\cos\left(\theta\right)\right)=0$
$\displaystyle \Rightarrow \sin^2\left(\theta\right)=\left(1+\cos\left(\theta \right)\right)\left(1-\cos\left(\theta\right)\right)$
$\displaystyle \Rightarrow \sin^2\left(\theta\right)=1-\cos^2\left(\theta\right)$
$\displaystyle \Rightarrow \sin^2\left(\theta\right)+\cos^2\left(\theta\right )=1$

which is the Pythagorean identity.
This method, is incorrect. You no longer have an identity if you are adjusting BOTH sides of the equation. Technically the original equation is this:

$\displaystyle \frac{\sin \Theta }{1-\cos \Theta }-\frac{1+\cos \Theta }{\sin \Theta }\Leftrightarrow0$

Is there an explanation of what i did...Like if I were to explain this problem.....any help for a step by step explanation?
Your goal, is to transform one side of your equation into the other side using Algebra, and already proven Trigonometric identities. As the RHS of the implication is just "0", there are any number of ways someone could come up with that, and you would have to be quite clever to produce something that gets you to the LHS (its not impossible, as you know the top must resolve to 0, but its not something youre meant to do).

Therefore your only options are to start working on the LHS. From the first blush, I see that there isn't a whole lot I can do with this equation. You could certainly break up the second part of the LHS into cosecants and tangents, but that doesn't present a clear path. However, once you start cross-multiplying you start seeing that the top is going to give you some further identities to work with (namely the Pythagorean identity), that you can use to get rid of some sines and cosines. Infact, you should note, that once you cross multiply and get common denominators, that your top will resolve to "0" and you are done with your solving.

9. Originally Posted by ANDS!
This method, is incorrect. You no longer have an identity if you are adjusting BOTH sides of the equation. Technically the original equation is this:

$\displaystyle \frac{\sin \Theta }{1-\cos \Theta }-\frac{1+\cos \Theta }{\sin \Theta }\Leftrightarrow0$
Yes, I was wrong.