Question

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

Answer 1

Verify it

Answer 2

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

Answer 3

I think that you have to prove it.

Answer 4

yeahh..

Answer 5

wait verify or prove?

Answer 6

suppose to verify the identity alegebraically

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

like i just did

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

wait, cot is negative??

Answer 17

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

Answer 18

ohright okay

Answer 19

lol cosß

Answer 20

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

Answer 21

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

Answer 22

i think that its wrong somewhere :/

Answer 23

I got it figured already though ;D

Answer 24

Lols, im way too slow~ xD

Answer 25

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

Answer 26

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

Answer 27

aww... okii, one moment..

Answer 28

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

Answer 29

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