remember from trig identity that sin²θ + cos²θ = 1
So, 1 - sin²θ = cos²θ
and 1- cos²θ = sin²θ
(1-sin²θ)/(1-cos²θ) = cos²θ / sin²θ = cot²θ
Cot^2x
(1 - sin^2)/(1 - cos^2)
(1 - sin^2/sin^2
1/sin^2 - 1
csc^2 - 1
tan^2
(1-sin²θ)/(1-cos²θ) =
cos²θ/sin²θ =
cot²θ
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Verified answer
remember from trig identity that sin²θ + cos²θ = 1
So, 1 - sin²θ = cos²θ
and 1- cos²θ = sin²θ
(1-sin²θ)/(1-cos²θ) = cos²θ / sin²θ = cot²θ
Cot^2x
(1 - sin^2)/(1 - cos^2)
(1 - sin^2/sin^2
1/sin^2 - 1
csc^2 - 1
tan^2
(1-sin²θ)/(1-cos²θ) =
cos²θ/sin²θ =
cot²θ