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Mathematics 19 Online
OpenStudy (anonymous):

Trying to solve a trig identity problem: sin x +1/ cos x + cot x.... I worked it almost to the end and my result was : sin x + sin ^2x / (sin x)(cos x) + cos x... Can any one help me?

OpenStudy (anonymous):

Where is the = sign?

OpenStudy (anonymous):

oh sorry i forgot... = tan sin x + 1 / cos x + cot x = tan x

OpenStudy (anonymous):

What did you do first?

OpenStudy (anonymous):

The first step I did was change the cot in denominator into cos and sin

OpenStudy (anonymous):

After I did that then I precedded with 1+ sin x/ (sin x)(cos x) + cos x/ sin

OpenStudy (anonymous):

Tell me in the original problem. is sin x by itself and then 1/cosx by itself and then cotx by itself....

OpenStudy (anonymous):

Ok, the original problem : (sin x +1) / (cos x + cot x) = tan x

OpenStudy (anonymous):

Oh.. ok then just a minute.

OpenStudy (anonymous):

Here is what to do.. let me know if you are following me. In algebra when you have a denominator and you are solving you usually multiply by the denominator so you can get rid of the denominator. So. I am going to multiply both sides by (cos x + cot x).. If you do you get (sin x + 1) = tan x(cos x + cot x) OK?

OpenStudy (anonymous):

I don't follow... i thoght you multiply by the reciprocal..

OpenStudy (anonymous):

i have to work on one side to equal the other side, which is tan x

OpenStudy (anonymous):

(cos x + cot x) times (sin x + 1) /(cos x + cot x) = tan x (cos x + cot x) The (cos x + cot x) would cancel out on the left hand side. It is legal to multiply both sides by the same thing.

OpenStudy (anonymous):

In math, you don't have to just work on one side. If you want to do it your way. give me a second. It is very easy if you do it my way... but just a second and I will get back with you shortly.

OpenStudy (anonymous):

Yes, it is legal to do both sides, however my professor wants us to work it by choosing complex side to prove other side

OpenStudy (anonymous):

It is WAY to messy to try and even get it your way.... if you had x/3 + 1/5 = 10 You should have been taught to get the common denominator of 15 and multiply every term by 15 which would give you 15 * (x/3) + 15 * (1/5) = 10 * 15 Do you remember doing this? It gets rid of the denominators for you.. you would have 5x + 3 = 150

OpenStudy (anonymous):

OK. just a minute.

OpenStudy (anonymous):

Yes, I remember doing this.

OpenStudy (anonymous):

OK.. Here is how you do it. going to be hard to write but here goes. (sin x + 1) / (cos x + (cos x/sin x)) = tan x { I substituted cos x / sinx for cot x OK so far?

OpenStudy (anonymous):

I understand that part

OpenStudy (anonymous):

Step 2 ... only look at the denominator and get a common denominator so cos x + (cos x/sin x) would have a common denominator od sinx x.. so I need to multiply cos x by (sin x / sin x) to get that common denominator. You would have (cos x sin x)/sinx + (cos x/sinx) on the bottom which turns into (cos x sin x) + cos x ALL OVER sin x OK?

OpenStudy (anonymous):

sorry common denominator of sin x

OpenStudy (anonymous):

ok, I got that too

OpenStudy (anonymous):

OK. almost done When we take a fraction divided by a fraction we multiply by the reciprocal.. so we have (sin x + 1) divided by (cos x sin x + cos x)/sin x So we would multiply by the reciprocal to give us sin x + 1 times sin x /(cos x sin x + cos x) OK?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Now you have to factor the denominator cos x sin x + cos x we factor that by taking out a common factor ex. 3x + 3 would factor into 3(x + 1) so this would be cos x(sin x + 1) OK?

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

So then you would have (sin x + 1) * sin x/(cos x(sin x + 1)) If you write this as a fraction you will see that (sin x + 1) cancels out You are then left with sin x / cos x Which we know that sin x/cos x = tan x SO tan x = tan x

OpenStudy (anonymous):

yes, thank you very much for showing the steps, it helps greatly, i started it but got stuck in the middle, thank you very much!

OpenStudy (anonymous):

You are welcome

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