Establish the identity. (cotθ+ cscθ+ 1)/(cotθ- cscθ- 1) =...

Establish the identity. \frac{cot\theta + csc\theta + 1}{cot\theta - csc\theta - 1} = -cot\theta - csc\theta To establish the identity, start with the side containing the more complicated expression. Here, expressions on both sides are of the same level of complexity, so start with the expression on the right side. - cot\theta - csc\theta = = = \frac{cot\theta + csc\theta + 1}{cot\theta - csc\theta - 1}

Transcript

00:02 All right, we are proving in this problem that the cosecant of theta minus 1 over the cotangent of theta is equal to the cotangent of theta over the cocecant of theta plus 1.

00:32 Now, in this particular problem, i don't know that one side is better to use than the other.

00:39 They look to be about the same.

00:40 I'm going to go ahead and work on the left side here and try to show that that's the same as the right side.

00:48 Now, there's not a lot of obvious things we can do.

00:54 One method we could follow right now is to get everything written in terms of sign and cosine.

00:59 That's always a good method that tends to work.

01:02 There's a shorter way to do this particular problem.

01:05 The way i would do this problem would be to recognize right now that i have, in my denominator on the right, i have the cosecane of theta plus one.

01:18 So i somehow need to get cosecant of theta plus 1 in my denominator on the left side.

01:25 So the first thing i'm going to do to start working on this proof then is i'm going to multiply this proof, or this expression.

01:34 I'm going to multiply the left side of it by the co -sicant of theta plus 1 over the cosecant of theta plus 1.

01:43 See, we really can do anything we want in here as long as we're performing legal math operations.

01:49 And technically speaking right there, i'm only multiplying by one, right? whenever you multiply something by, multiply by something over itself, you're multiplying by one.

01:59 What this will allow us to do is when i multiply my numerators and i distribute, cosecant theta times co -secant theta will be the co -secant squared of theta...

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