Asked by VD

(csc(B)+cot (B))(csc(B)-cot(B))
Answered by Reiny
You have the pattern of "difference of squares"

(csc(B)+cot (B))(csc(B)-cot(B))
= csc^2 B - cot^2 B

But csc^ B = 1 + cot^2B , one of the identies from
sin^2 B + cos^2 B= 1

so csc^2 B - cot^2 B
= 1 + cot^2 B - cot^2 B = 1
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