OpenStudy (anonymous):

\[\frac{1+\csc \theta}{\sec \theta} -\cot \theta = \cos \theta\]

5 years ago
OpenStudy (anonymous):

Verify it

5 years ago
OpenStudy (anonymous):

just plug in numbers to verify or graph the two functions and they should overlap each otheres

5 years ago
OpenStudy (mimi_x3):

I think that you have to prove it.

5 years ago
OpenStudy (anonymous):

yeahh..

5 years ago
OpenStudy (anonymous):

wait verify or prove?

5 years ago
OpenStudy (anonymous):

suppose to verify the identity alegebraically

5 years ago
OpenStudy (anonymous):

then i guess you need to prove it. First converts everything in terms of sine and cosine

5 years ago
OpenStudy (anonymous):

okay, I'll show where I got to, and maybe you can help me figure out what I did wrong @@

5 years ago
myininaya (myininaya):

\[\frac{1+\frac{1}{\sin(\theta)}}{\frac{1}\cos(\theta)}\]

5 years ago
myininaya (myininaya):

the first thing to do is write in terms of sin and cos

5 years ago
myininaya (myininaya):

like i just did

5 years ago
myininaya (myininaya):

now multiply top and bottom by cos to get rid of the compound fraction

5 years ago
myininaya (myininaya):

\[\frac{1+\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}} \cdot \frac{\cos(\theta)}{\cos(\theta)}\]

5 years ago
myininaya (myininaya):

\[\frac{\cos(\theta)+\frac{\cos(\theta)}{\sin(\theta)}}{1}=\cos(\theta)+\cot(\theta)\]

5 years ago
myininaya (myininaya):

now we still have that -cot(theta) to worry about

5 years ago
OpenStudy (anonymous):

wait, cot is negative??

5 years ago
myininaya (myininaya):

guess what this equals \[\cos(\theta)+\cot(\theta)-\cot(\theta)\]

5 years ago
OpenStudy (anonymous):

ohright okay

5 years ago
OpenStudy (anonymous):

lol cosß

5 years ago
OpenStudy (anonymous):

@myiniaya these problems keep getting bigger ;c now they have natural logs...

5 years ago
OpenStudy (mimi_x3):

\[\large \frac{1+\frac{1}{sinx}}{\frac{1}{cosx}} - \frac{cosx}{sinx}= \frac{sinx+1}{\frac{sinx}{\frac{1}{cosx}}} - \frac{cosx}{sinx} =\] \[\large \frac{sinx+1}{sinx} *\frac{cosx}{1} - \frac{cosx}{sinx} = \frac{sinxcosx+cosx}{sinx} - \frac{cosx}{sinx} \] \[\frac{sinxcosx+cosx-cosx}{sinx} = \frac{sinxcosx}{sinx} =cosx\]

5 years ago
OpenStudy (mimi_x3):

i think that its wrong somewhere :/

5 years ago
OpenStudy (anonymous):

I got it figured already though ;D

5 years ago
OpenStudy (mimi_x3):

Lols, im way too slow~ xD

5 years ago
OpenStudy (anonymous):

it's okii ;D You wanna work another with me? It has natural logs...

5 years ago
OpenStudy (mimi_x3):

Hmm..i will have to see what it is first, im tired now xD

5 years ago
OpenStudy (anonymous):

aww... okii, one moment..

5 years ago
OpenStudy (anonymous):

\[\ln \left| 1+\csc \theta \right|+\ln \left| 1-\cos \theta \right| =2\ln \left| \sin \theta \right|\]

5 years ago
OpenStudy (mimi_x3):

Sorry, too tired now. Post it someone else can help you.

5 years ago