help with trig identities

[rickb]

problem reads as follows:

1. (tan x)/ (1 + tan x)/ (cot x)/ (1 + tan x) = 1 - cot x

2. sec^2 x - csc^2 x= (tan x - cot x)/ (sin x)(cos x)

 

[galactus]

#2: It may be easier to convert the right into the left side.

\(\displaystyle \frac{tan(x)-cot(x)}{sin(x)cos(x)}\)

\(\displaystyle \frac{\frac{sin(x)}{cos(x)}-\frac{cos(x)}{sin(x)}}{sin(x)cos(x)}\)

\(\displaystyle \frac{sin^{2}(x)-cos^{2}(x)}{sin^{2}(x)cos^{2}(x)}\)

Now, can you finish?. A few more small steps and you have it.

 

[rickb]

this is where I am stuck

 

[stapel]

this is where I am stuck

It would have been helpful to have shown that you'd gotten this far. This is why the "Read Before Posting" thread that you read before posting specified that you "Show all of your work. If you've shown no work at all, the tutors may assume that you're needing help getting started, and may suggest only how to do the first step. Even if you're asking only about the very end of the solution process, still include all of your intermediate steps." Claiming to have been able to complete the difficult portion that the tutor did for you, while being unable to finish the easy part, tends to lead to the conclusion that the poster is "fishing" for the complete worked solution without actually doing any work of his own. Surely this was not the impression you'd meant to give. So:

To complete exercise (2), divide the fraction into its two terms, cancel each term's common factor, and convert from sines and cosines to secants and cosecants.

Please reply with a complete listing of what you have done so far for exercise (1).

Thank you.

Eliz.

 

[galactus]

\(\displaystyle \frac{sin^{2}(x)-cos^{2}(x)}{sin^{2}(x)cos^{2}(x)}\)

\(\displaystyle \frac{sin^{2}(x)}{sin^{2}(x)cos^{2}(x)}-\frac{cos^{2}(x)}{sin^{2}(x)cos^{2}(x)}\)

\(\displaystyle \frac{1}{cos^{2}(x)}-\frac{1}{sin^{2}(x)}\)

Now, see it?.

Try the other and let us know how it goes.

 

[rickb]

sorry, this is the first time I have used this forum, for the other problem I listed I have tried converting everthing on the left side of the equation to sin and cos and have had no luck

 

[stapel]

for the other problem I listed I have tried converting everthing on the left side of the equation to sin and cos and have had no luck

You might be nearly there! Please reply showing what you have so far. Thank you!

Eliz.

 

[rickb]

For the left side I have the following:

(sin x)/(cos x)/ 1+ (sin x/con x) - (cos x)/(sin x)/ 1+ (sin x/cos x)

I thought I could use the conjugate of the denominator but didn't have any luck

 

[Deleted member 4993]

For the left side I have the following:

(sin x)/(cos x)/ 1+ (sin x/con x) - (cos x)/(sin x)/ 1+ (sin x/cos x) <<<< (tan x)/ (1 + tan x)/ (cot x)/ (1 + tan x) does not match with what you wrote - which one is your actual LHS

I am assuming your actual problem is:

(tan x)/ (1 + tan x) - (cot x)/ (1 + tan x) = 1 - cot x

Since RHS is expressed in terms of 'cot x' only - try expressing LHS in terms of 'cot x' . The algebraic manipulations will be simpler.

(tan x)/ (1 + tan x) - (cot x)/ (1 + tan x)

(1/cot x)/( 1 + 1/cot x) - cot x/ (1 + 1/cot x)

= 1/(1 + cot x) - cot^2(x)/(1 + cot x)

Now clean up and you are done...

I thought I could use the conjugate of the denominator but didn't have any luck