How do you prove $(csc (t) - 1) (csc (t) + 1) = {cot}^{2} (t)$ $(csc(t)-1)(csc(t)+1)=cot^2(t)$ ?

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How do you prove $(csc (t) - 1) (csc (t) + 1) = {cot}^{2} (t)$ $(csc(t)-1)(csc(t)+1)=cot^2(t)$ ?

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Answer 1 · 2015-06-23T03:39:20

Prove: (csc x - 1)(csc x + 1) = cot^2x

Explanation:

$(\frac{1}{sin x} - 1) (\frac{1}{sin x} + 1) = (\frac{1}{{sin}^{2} x} - 1) = \frac{1 - {sin}^{2} x}{{sin}^{2} x}$ $(1/sin x - 1)(1/sin x + 1) = (1/sin^2 x - 1) = (1 - sin^2 x)/sin^2 x$ =

$= \frac{{cos}^{2} x}{{sin}^{2} x} = {cot}^{2} x$ $= cos^2 x/sin^2 x = cot^2 x$

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How do you prove $(csc (t) - 1) (csc (t) + 1) = {cot}^{2} (t)$ $(csc(t)-1)(csc(t)+1)=cot^2(t)$ ?

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How do you prove (csc(t)−1)(csc(t)+1)=cot2(t)(csc(t)-1)(csc(t)+1)=cot^2(t)?

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How do you prove $(csc (t) - 1) (csc (t) + 1) = {cot}^{2} (t)$ $(csc(t)-1)(csc(t)+1)=cot^2(t)$ ?