How do you verify the identity #(csctheta-cottheta)(csctheta+cottheta)=1#?

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Answer 1 · 2016-09-09T17:13:56

We have: #(csc(theta) - cot(theta)) (csc(theta) + cot(theta))#

Let's expand the parentheses:

#= (csc(theta)) (csc(theta)) + (csc(theta)) (cot(theta)) + ( - cot(theta) (csc(theta)) + (- cot(theta)) (cot(theta))#

#= csc^(2)(theta) + csc(theta) cot (theta) - csc(theta) cot(theta) - cot^(2)(theta)#

#csc^(2)(theta) - cot^(2)(theta)#

Then, let's apply two standard trigonometric identities; #csc(theta) = (1) / (sin(theta))# and #cot(theta) = (cos(theta)) / (sin(theta))#:

#= ((1) / (sin(theta)))^(2) - ((cos(theta)) / (sin(theta)))^(2)#

#= (1) / (sin^(2)(theta)) - (cos^(2)(theta)) / (sin^(2)(theta))#

#= (1 - cos^(2)(theta)) / (sin^(2)(theta))#

One of the Pythagorean identities is #cos^(2)(theta) + sin^(2)(theta) = 1#.

We can rearrange this to get:

#=> sin^(2)(theta) = 1 - cos^(2)(theta)#

Let's apply this rearranged identity to our proof:

#= (sin^(2)(theta)) / (sin^(2)(theta))#

#=1# (Q.E.D.)

Answer 2 · 2016-09-09T17:17:06

I changed all into #sin# and #cos#:

have a look:
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