Prove using Trigonometric Identities

**evanator** · October 24th 2010, 07:51 AM

Hi all,

I am having trouble with the following question. I have filled a few pages with various manipulations and I am no closer to proving the identity. I have obviously missed a trick. Any help would be appreciated.

First the question:

Prove that $\frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} \equiv \sin \theta + \cos \theta$

Let's take the right hand side and see what we can do with it:

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

$\frac{\sin^2 \theta}{\sin \theta} + \frac{\cos^2 \theta}{\cos \theta}$

$\frac{1 - \cos^2 \theta}{\sin \theta} + \frac{1 - \sin^2 \theta}{\cos \theta}$

$\frac{1}{\sin \theta} - \frac{\cos^2 \theta}{\sin \theta} + \frac{1}{\cos \theta} - \frac{\sin^2 \theta}{\cos \theta}$

$\frac{\sin \theta - \sin \theta \cos^2 \theta}{\sin^2 \theta} + \frac{\cos \theta - \cos \theta \sin^2 \theta}{\cos^2 \theta}$

$\frac{\sin \theta -\sin \theta \cos^2 \theta}{1-\cos^2 \theta} + \frac{\cos \theta - \cos \theta \sin^2 \theta}{1 - \sin^2 \theta}$

I can't really see where to go with it.

Regards,

Evanator

**skeeter** · October 24th 2010, 08:02 AM

Originally Posted by evanator

Hi all,

I am having trouble with the following question. I have filled a few pages with various manipulations and I am no closer to proving the identity. I have obviously missed a trick. Any help would be appreciated.

First the question:

Prove that $\frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} \equiv \sin \theta + \cos \theta$

$\displaystyle \frac{\cos{t}}{1-\tan{t}} \cdot \frac{\cos{t}}{\cos{t}} + \frac{\sin{t}}{1-\cot{t}} \cdot \frac{\sin{t}}{\sin{t}} =$

$\displaystyle \frac{\cos^2{t}}{\cos{t} - \sin{t}} + \frac{\sin^2{t}}{\sin{t} - \cos{t}} =$

$\displaystyle \frac{\cos^2{t} - \sin^2{t}}{\cos{t} - \sin{t}} =$

last two steps I leave for you ...

**Unknown008** · October 24th 2010, 08:05 AM

It is usually much more easier to go from 'big ones' to 'smaller ones', that is the one on the left.

$\begin{array}{lcl} \dfrac{\cos \theta}{1 - \tan \theta} + \dfrac{\sin \theta}{1 - \cot \theta} &=& \dfrac{\cos \theta}{1 - \frac{\sin\theta}{\cos\theta}} + \dfrac{\sin \theta}{1 - \frac{\cos\theta}{\sin\theta}} \\ && \\ &=& \dfrac{\cos \theta}{ \frac{\cos\theta - \sin\theta}{\cos\theta}} + \dfrac{\sin \theta}{\frac{sin\theta - \cos\theta}{\sin\theta}} \\ && \\ &=& \dfrac{\cos^2 \theta}{\cos\theta - \sin\theta} + \dfrac{\sin^2 \theta}{\sin\theta - \cos\theta} \\ && \\ &=& \dfrac{\cos^2 \theta}{\cos\theta - \sin\theta} - \dfrac{\sin^2 \theta}{\cos\theta - \sin\theta} \\ && \\ &=& \dfrac{\cos^2 \theta - \sin^2\theta}{\cos\theta - \sin\theta}\end{array} \end{array}$

Can you complete it now?

EDIT: A tad too late

**evanator** · October 24th 2010, 03:24 PM

$\frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta - \sin \theta} =$

$\frac{(\cos \theta + \sin \theta)(\cos \theta - \sin \theta)}{\cos \theta - \sin \theta} =$

$\cos \theta + \sin \theta$

Thank you both, and thank you to whoever moved my thread. I'm always quite annoyed when I see how straightforward it is, but I'm sure I'll get a feel for it with time.

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