C2 another trigonometrical equation

Hi, still on the trigonometrical equations…finding them quite difficult to be honest. Got this one here that I’m not sure where to begin with:

$\frac {\sin \theta}{1 \cos \theta} + \frac {1 \cos \theta}{\sin \theta} \equiv \frac {2}{\sin \theta}$

I’m really not being lazy but I have no idea where to begin with, I didn’t think the $\tan \theta \equiv \frac {\sin \theta}{\cos \theta}$ identity would be useful as adding tan into the identity would just cause another variable, and I also didn’t think it would be the $\sin^2 \theta + \cos^2 \theta \equiv 1$ identity as there are no squared values, so I’m lost!

Is your question, $\displaystyle \frac{sin\theta}{cos\theta}+ \frac{cos\theta}{sin\theta} \equiv \frac2{sin2\theta} \ ?$

You want to rearrange the equation to get it into a useful form to start of with, see what you can do and see if you can use a trig identity after rearranging.

EDIT: If you want the solution I have it if you are really stuck.

It is not an equation. I think the question is show LHS equals RHS

OK $\frac {\sin \theta}{1 \cos \theta} + \frac {1 \cos \theta}{\sin \theta}$ so if I try to sort out the numerators and denominators out they just result in 1, don't they?

$\frac {\sin \theta + 1 \cos \theta }{1 \cos \theta + \sin \theta }$ doesn’t this just result in 1?

You're adding them wrong. Remember:

$\frac{a}{b} + \frac{c}{d}= \frac{ad+bc}{bd}$

It's an identity.




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It's an identity.




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Doesn't seem like OP's question is valid, I think his question is to prove 2/sin2(theta)

The question copied from my book, word for word is:

Prove the identity $\frac {\sin \theta}{1 \cos \theta} + \frac {1 \cos \theta}{\sin \theta} \equiv \frac {2}{\sin \theta}$

that's incorrect...