Please check proof of identity.

**Furyan** · December 27th 2011, 05:07 PM

Hello

This is the last identity, in this exercise.

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

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 $1 - \sin\theta$ and $\sec\theta + 1$ gets me to:

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

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

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

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

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

QED?

Thank you

**MarkFL** · December 27th 2011, 05:55 PM

Looks good to me!

**Siron** · December 27th 2011, 10:50 PM

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

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

**Furyan** · December 28th 2011, 03:16 AM

Hi Siron

Originally Posted by Siron

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

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

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.

**Siron** · December 28th 2011, 03:29 AM

You're welcome!

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

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