Basic identity proving

**Tyln** · April 5th 2009, 01:10 PM

Hi,

I just started with trig and trig proofs are like really hard for me, I just can't seem to understand them. This is the first question in my workbook..

$\frac{\cot \theta -1}{\tan \theta -1}= -\frac{1}{\tan \theta}$

it asks me to "Verify the statement is true for $\theta=\frac{\pi}{3}$ " and then "Prove the statement is an identity." Could someone please show me the steps on doing this? The right side I'm actually okay with but the left side, I don't know what to do with the "- 1s"

Oh, and any tips on how I can learn how to do these types of questions and eventually advance in trig proving, they would be really appreciated!

**Jhevon** · April 5th 2009, 01:23 PM

Originally Posted by Tyln

Hi,

I just started with trig and trig proofs are like really hard for me, I just can't seem to understand them. This is the first question in my workbook..

$\frac{\cot \theta -1}{\tan \theta -1}= -\frac{1}{\tan \theta}$

it asks me to "Verify the statement is true for $\theta=\frac{\pi}{3}$ " and then "Prove the statement is an identity." Could someone please show me the steps on doing this? The right side I'm actually okay with but the left side, I don't know what to do with the "- 1s"

verify for $\theta = \frac {\pi}3$ means to plug in $\theta = \frac {\pi}3$ and show that both sides of the equation gives the same thing. that being said, this is not a complete identity. $\theta = \frac {\pi}4$ works on the right, but not on the left.

anyway, to verufy the "identity", note that the right hand side is $- \frac {\cos x}{\sin x}$

Now consider the LHS: write all the funcitons in terms of sines and cosines, simplify and show you get the expression above

Oh, and any tips on how I can learn how to do these types of questions and eventually advance in trig proving, they would be really appreciated!

TIPS:

- in general you want to start with one side and try to manipulate it to get the other side. to this end, it is usually best to start with the most complicated side as it gives you more options to change things.

- sometimes you will not be able, or will not "see" how, to manipulate one side to get to the other. in this case, simplify each side one at a time and show that you can bring both expressions to the same thing.

- if you get stuck, changing all functions to be in terms of sines and cosines is a good way to proceed. these are the trig functions students are most familiar with, and so it is easier to spot relevant manipulations if you are using these functions

**stapel** · April 5th 2009, 01:26 PM

Originally Posted by Tyln

$\frac{\cot \theta -1}{\tan \theta -1}= -\frac{1}{\tan \theta}$

It's usually a good idea to start with the more-complicated side, and it's often helpful to convert everything to sines and cosines, and to find common denominators, as needed.

. . . . . $\frac{\cot(\theta)\, -\, 1}{\tan(\theta)\, -\, 1}\, =\, \frac{\frac{\cos(\theta)}{\sin(\theta)}\, -\, 1}{\frac{\sin(\theta)}{\cos(\theta)}\, -\, 1}$

. . . . . $=\, \frac{\left(\frac{\cos(\theta)\, -\, \sin(\theta)}{\sin(\theta)}\right)}{\left(\frac{\s in(\theta)\, -\, \cos(\theta)}{\cos(\theta)}\right)}$

. . . . . $=\, \frac{-\left(\frac{\sin(\theta)\, -\,\cos(\theta)}{\sin(\theta)}\right)}{\left(\frac{ \sin(\theta)\, -\, \cos(\theta)}{\cos(\theta)}\right)}$

Do the cancellation, and then simplify what's left.

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