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Question

$cos θ \sqrt{1 + {cot}^{2} θ} =$
$tan θ$
$cot θ =$
$sin θ =$
$cos θ$

A

tan θ

B

cos θ

C

cot θ =

D

sin θ =

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Solution

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Use 1 + ${cot}^{2} θ = c o s e c^{2} θ$

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