Question

Simplify; csc(theta)cot(theta)/sec(theta)

Answer 1

Using confuction identites would be the proper way to go about this, right? I'm not sure what to do after that (potential) fact though.

Answer 2

\[\frac{ \csc(\theta)*\cot(\theta) }{ \sec(\theta) }\] first identify what csc(theta), cot(theta) and sec(theta) are using their identities. So we know \[\sec(\theta)=\frac{ 1 }{ \cos(\theta) }, \csc(\theta)=\frac{ 1 }{ \sin(\theta) }, \cot(\theta)=\frac{ \cos(\theta) }{ \sin(\theta) }\] Now simply plug in all these givens. \[\large \frac{ \csc(\theta)*\cot(\theta) }{ \sec(\theta) }= \frac{ 1 }{\sin(\theta) }*\frac{ \cos(\theta) }{ \sin(\theta) }*\cos(\theta)\]

Answer 3

you're dividing by sec(theta) = 1/cos(theta) so that's like saying \[\large (\frac{ 1 }{ a }*\frac{ 1 }{ b })/\frac{ 1 }{ c }= (\frac{ 1 }{ a }*\frac{ 1 }{ b }) *c\]

Answer 4

Will the sin(theta) still cancel out if both are in the denominator?

Answer 5

but then tell me how the sin(theta) cancel out if you have them BOTH in the denominator. Wouldn't they just multiply to eachother?

Answer 6

Yes, I believe so. If that were the case, wouldn't it then become cos^2(theta)/sin^2(theta)?

Answer 7

Yes! :)

Answer 8

which is cot^2(theta)?

Answer 9

mmhmm :)

Answer 10

=D Thanks so much!

Answer 11

And then there is an identity for cot^(2) (theta) which is \[1+\cot ^{2}(\theta)= \csc ^{2}(\theta)\] and solving for cot^(2)(theta) we have \[\cot ^{2}(\theta)=\csc ^{2}(\theta)-1\] but this is just for further understanding i believe :) Like if you were given values to substitute in...

Answer 12

Yeah lol sometimes I wonder how I was even confused by an equation like this once I understand.

Answer 13

We all get like that sometimes :P And then we either a) learn from our mistakes or b) learn from other people's understanding of a problem :)