How do you prove $(1 - {cos}^{2} x) (1 + {cot}^{2} x) = 1$ $(1-\cos^2 x)(1+\cot^2 x) = 1$ ?

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Answer 1 · 2014-10-24T09:43:54

By the trig identities

${cos}^{2} x + {sin}^{2} x = 1 \Rightarrow {sin}^{2} x = 1 - {cos}^{2} x$ $cos^2x+sin^2x=1 Rightarrow sin^2x=1-cos^2x$

and

$1 + {cot}^{2} x = {csc}^{2} x = \frac{1}{{sin}^{2} x}$ $1+cot^2x=csc^2x=1/{sin^2x}$ ,

we have

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

I hope that this was helpful.

How do you prove (1-\cos^2 x)(1+\cot^2 x) = 1(1−cos2x)(1+cot2x)=1(1-\cos^2 x)(1+\cot^2 x) = 1?