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:
enter image source here

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