Please check proof of identity.

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

Re: Please check proof of identity.

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:

$\displaystyle \cot^2(x)[\sec(x)-1][1+\sec(x)]=\sec^2(x)[1-\sin(x)][1+\sin(x)]$

$\displaystyle \Leftrightarrow \cot^2(x)[\sec^2(x)-1]=\sec^2(x)[1-\sin^2(x)]$

$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1}{\cos^2(x)}-1\right]=\sec^2(x)\cos^2(x)$

$\displaystyle \Leftrightarrow \cot^2(x)\left[\frac{1-\cos^2(x)}{\cos^2(x)}\right]=\frac{\cos^2(x)}{\cos^2(x)}$

$\displaystyle \Leftrightarrow \cot^2(x)\frac{\sin^2(x)}{\cos^2(x)}=1$

$\displaystyle \Leftrightarrow \cot^2(x)\tan^2(x)=1$

$\displaystyle \Leftrightarrow 1=1$

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.

Re: Please check proof of identity.

You're welcome! ;)

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