((csc x+1)(csc x-1))/(csc^2x)\\ cscθ·cos^2θ+sinθ= cscθ

Question

`((csc x+1)(csc x-1))/(csc^2x)\\`

cscθ·cos^2θ+sinθ= cscθ

Answer 1

Explanation:

`(( \csc x+1)( \csc x-1))/( \csc ^2x) = \frac{ { \csc }^(2) x - 1}{ \csc^(2) x} \\ = 1 - \sin^(2) x = \cos^(2) x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:`

b. Since csc theta = 1/sin theta, we can multiply both sides by sin theta and you will end up with

`\cos^(2) \theta + \sin^(2) \theta = 1`

which is an identity.