What is $cot^2theta$ in terms of non-exponential trigonometric functions?

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Answer 1 · 2016-05-24T05:12:37

$(1 + cos 2t)/(1 - cos 2t)$

$cot^2 t = (cos^2 t)/(sin^2 t)$
$cos^2 t = (1/2)(1 + cos 2t)$
$sin^2 t = (1/2)(1 - cos 2t)$
Therefor,
$cot^2 t = (1 + cos 2t)/(1 - cos 2t)$

What is cot^2thetacot^2theta in terms of non-exponential trigonometric functions?