trig identity: sin @ / (1-cos @) - (1 + cos @) / sin @ = 0

[sonyanale]

Please help me to prove this mathematical identity.
sin theta / (1- cos theta) - (1+ cos theta)/sin theta=0

 

[galactus]

Re: Math identity --Please help!

Cross multiply. The numerator becomes $\displaystyle sin^{2}(t)-(1-cos^{2}(t))$.

See what that is?.

 

[soroban]

Re: Math identity --Please help!

Hello, sonyanale!

Another approach . . .

$\displaystyle \text{Prove: }\:\frac{\sin\theta}{1- \cos\theta} - \frac{1+ \cos\theta}{\sin\theta} \:=\:0$

$\displaystyle \text{Multiply the first fraction by }\,\frac{1+\cos\theta}{1+\cos\theta}$

$\displaystyle \frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{1 - \cos^2\!\theta} \;=\;\frac{\sin\theta(1+\cos\theta)}{\sin^2\!\theta} \;=\;\frac{1+\cos\theta}{\sin\theta}$


\(\displaystyle \text{And the problem becomes: }\:\frac{1+\cos\theta}{\sin\theta} - \frac{1+\cos\theta}{\sin\theta} \quad\hdots\;\text{ which equals 0.}\)

 

[sonyanale]

Re: Math identity --Please help!

Please elaborate with each step explained as in a proof. I am lost. please help.

 

[galactus]

Sonya, with respect, if you're that lost you need to see your instructor. Soroban gave about as much detail as can be shown. I just showed a simple identity

$\displaystyle 1-cos^{2}(t)=sin^{2}(t)$.

That leaves 0 in the numerator and it's finished.

$\displaystyle sin^{2}(t)-(1-cos^{2}(t))=0$

 

[sonyanale]

I thought you did not know what you were doing when you said to cross multiply because the book says not to assume that both sides are equal. It also says that you cannot prove an identity by first assuming that both sides are equal. Sorry. I just wanted more explanations with each step on the other reply.

Thanks anyway

 

[galactus]

I did not assume they were equal when I cross multiplied the left side. I did not cross multiply across the equals sign.