# Please check proof of identity.

• Dec 27th 2011, 05:07 PM
Furyan
Hello

This is the last identity, in this exercise.

$\displaystyle \dfrac{\cot^2\theta(\sec\theta\ – 1)}{1 + \sin\theta} \equiv \dfrac{\sec^2\theat(1 - \sin\theta)}{1 + \sec\theta}$

I seem to have proved it, but I would appreciate someone checking my work to make sure that I haven’t got there by accident.

Multiplying the numerator and denominator by $\displaystyle 1 - \sin\theta$ and $\displaystyle \sec\theta + 1$ gets me to:

$\displaystyle \dfrac{\cot^2\theta(\sec^2\theta\ – 1)(1 - \sin\theta)}{\cos^2\theta(\sec\theta + 1)}$

$\displaystyle \cot^2\theta(\sec^2\theta\ – 1) = 1$, which gives:

$\displaystyle \dfrac{1 - \sin\theta}{\cos^2\theta(\sec\theta + 1)}$

Multiplying numerator and denominator by $\displaystyle \sec^2\theta$ gives:

$\displaystyle \dfrac{\sec^2\theat(1 - \sin\theta)}{1 + \sec\theta}$

QED?

Thank you
• Dec 27th 2011, 05:55 PM
MarkFL
Re: Please check proof of identity.
Looks good to me! :)
• Dec 27th 2011, 10:50 PM
Siron
Re: Please check proof of identity.
You can also prove the identity by using:
$\displaystyle \frac{A}{B }=\frac{C}{D} \Leftrightarrow A\cdot D=B \cdot C$

That means we have to prove:
$\displaystyle \cot^2(x)[\sec(x)-1][1+\sec(x)]=\sec^2(x)[1-\sin(x)][1+\sin(x)]$
$\displaystyle \Leftrightarrow \cot^2(x)[\sec^2(x)-1]=\sec^2(x)[1-\sin^2(x)]$
$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1}{\cos^2(x)}-1\right]=\sec^2(x)\cos^2(x)$
$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1-\cos^2(x)}{\cos^2(x)}\right]=\frac{\cos^2(x)}{\cos^2(x)}$
$\displaystyle \Leftrightarrow \cot^2(x)\frac{\sin^2(x)}{\cos^2(x)}=1$
$\displaystyle \Leftrightarrow \cot^2(x)\tan^2(x)=1$
$\displaystyle \Leftrightarrow 1=1$
• Dec 28th 2011, 03:16 AM
Furyan
Re: Please check proof of identity.
Hi Siron :)

Quote:

Originally Posted by Siron
You can also prove the identity by using:
$\displaystyle \frac{A}{B }=\frac{C}{D} \Leftrightarrow A\cdot D=B \cdot C$

That means we have to prove:
$\displaystyle \cot^2(x)[\sec(x)-1][1+\sec(x)]=\sec^2(x)[1-\sin(x)][1+\sin(x)]$
$\displaystyle \Leftrightarrow \cot^2(x)[\sec^2(x)-1]=\sec^2(x)[1-\sin^2(x)]$
$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1}{\cos^2(x)}-1\right]=\sec^2(x)\cos^2(x)$
$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1-\cos^2(x)}{\cos^2(x)}\right]=\frac{\cos^2(x)}{\cos^2(x)}$
$\displaystyle \Leftrightarrow \cot^2(x)\frac{\sin^2(x)}{\cos^2(x)}=1$
$\displaystyle \Leftrightarrow \cot^2(x)\tan^2(x)=1$
$\displaystyle \Leftrightarrow 1=1$

I really like that method, a lot. It seems more natural to me than trying to manipulate one side into the other. Thank you very much for taking the time to post it. I've got more identities to prove in other chapters and I'll be looking to use your method whenever I can.

Thanks again. It's enormously helpful, there's nothing like that in any of my books.
• Dec 28th 2011, 03:29 AM
Siron
Re: Please check proof of identity.
You're welcome! ;)

Like you see, it can be useful to prove rational identities.