1. ## Proving identities 2

Hello, me again

I'm stuck on another identity:

$\dfrac{1 - \sin\theta}{\cos\theta} \equiv \dfrac{1}{\sec\theta + \tan\theta}$

I got:

$LHS \equiv \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta} \equiv \sec\theta - \tan\theta$

I tried starting with RHS and couldn’t do it that way either. I'd really appreciate some help.

Thank you

2. ## Re: Proving identities 2

Originally Posted by Furyan
$\dfrac{1 - \sin\theta}{\cos\theta} \equiv \dfrac{1}{\sec\theta + \tan(\theta)}$
Change the RHS to $\frac{\cos(\theta)}{1+\sin(\theta)}$
Then multiply by $\frac{1-\sin(\theta)}{1-\sin(\theta)}$.

3. ## Re: Proving identities 2

Originally Posted by Furyan
Hello, me again

I'm stuck on another identity:

$\dfrac{1 - \sin\theta}{\cos\theta} \equiv \dfrac{1}{\sec\theta + \tan\theta}$

I got:

$LHS \equiv \dfrac{1}{\cos\theta} - \dfrac{\sin\theta}{\cos\theta} \equiv \sec\theta - \tan\theta$

I tried starting with RHS and couldn’t do it that way either. I'd really appreciate some help.

Thank you
That works. Next multiply numerator and denominator by $\sec \theta + \tan \theta$ to give $\dfrac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\sec \theta + \tan \theta}$

You can simplify that numerator using the difference of two squares (this is a nifty little formula!) which in turn can have an identity used on it - try rearranging $\tan^2 \theta + 1 = \sec^2 \theta$ if you get stuck

4. ## Re: Proving identities 2

Originally Posted by Plato
Change the RHS to $\frac{\cos(\theta)}{1+\sin(\theta)}$
Then multiply by $\frac{1-\sin(\theta)}{1-\sin(\theta)}$.
Thank you Plato,

That worked like a charm, really amazing. Can I ask did you choose $\dfrac{1 - \sin\theta}{1 - \sin\theta}$ because you could see it would give you $\cos^2\theta$ in the denominator? I haven't used a manipulation like that in any other question I've done, but I have a feeling if I had they may have been very much easier. I have a similar one I was stuck on I'll try it on that and see if it helps.

Thanks again.

5. ## Re: Proving identities 2

Originally Posted by Furyan
Can I ask did you choose $\dfrac{1 - \sin\theta}{1 - \sin\theta}$ because you could see it would give you $\cos^2\theta$ in the denominator?
To be truthful, over fifty years ago I won a citywide identities proving contest. I just never lost the ability to see the needed trick.

6. ## Re: Proving identities 2

Originally Posted by e^(i*pi)
That works. Next multiply numerator and denominator by $\sec \theta + \tan \theta$ to give $\dfrac{(\sec \theta - \tan \theta)(\sec \theta + \tan \theta)}{\sec \theta + \tan \theta}$

You can simplify that numerator using the difference of two squares (this is a nifty little formula!) which in turn can have an identity used on it - try rearranging $\tan^2 \theta + 1 = \sec^2 \theta$ if you get stuck
Thank you e^i*pi, that worked like a charm too. Thanks for the hint about $\sec^2\theta - \tan^2\theta \equiv 1$ I hadn't used that and might not have seen it. Now I know I should try using that equivalent fraction technique if I'm really stuck, which I have been on another on for some time. I hope it works on it. The difference of two squares has come up a lot. Plato's method above uses it in the denominator.

Thanks again

7. ## Re: Proving identities 2

Originally Posted by Plato
To be truthful, over fifty years ago I won a citywide identities proving contest. I just never lost the ability to see the needed trick.
Thanks, that explains how you can solve in a micro second something I've been working on for hours. I was working on others that I could do in that time too though.