Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
|
One Answer Below
Anyone can contribute an answer, even non-tutors.
Idk
Date Posted: 3 years ago
Views: 697
1 - cot2θ tanθ
= (1/tanθ - cot2θ) (tanθ)
= (cosθ/sinθ - cos2θ/sin2θ) (tanθ)
= (cosθ/sinθ - (cos²θ - sin²θ)/(2sinθcosθ) ) (tanθ)
= ( 2cos²θ/(2sinθcosθ) + (sin²θ - cos²θ)/(2sinθcosθ) ) (tanθ)
= ((cos²θ+sin²θ)/2sinθcosθ) (tanθ)
= (1/sin2θ) (tanθ)
= cosec2θ tanθ
(Shown)
= (1/tanθ - cot2θ) (tanθ)
= (cosθ/sinθ - cos2θ/sin2θ) (tanθ)
= (cosθ/sinθ - (cos²θ - sin²θ)/(2sinθcosθ) ) (tanθ)
= ( 2cos²θ/(2sinθcosθ) + (sin²θ - cos²θ)/(2sinθcosθ) ) (tanθ)
= ((cos²θ+sin²θ)/2sinθcosθ) (tanθ)
= (1/sin2θ) (tanθ)
= cosec2θ tanθ
(Shown)
Alternatively,
1 - cot2θtanθ
= 1 - (1/tan2θ)(tanθ)
= 1 - ((1-tan²θ)/(2tanθ)) (tanθ)
= 1 - (1-tan²θ)/2
= ½(1 + tan²θ)
= ½sec²θ
= (1/2cosθ)(1/cosθ)
= (1/2cosθ)(1/sinθ)(sinθ)(1/cosθ)
= (1/2sinθcosθ)(sinθ/cosθ)
= (1/sin2θ) tanθ
= cosec2θ tanθ
(Shown)
1 - cot2θtanθ
= 1 - (1/tan2θ)(tanθ)
= 1 - ((1-tan²θ)/(2tanθ)) (tanθ)
= 1 - (1-tan²θ)/2
= ½(1 + tan²θ)
= ½sec²θ
= (1/2cosθ)(1/cosθ)
= (1/2cosθ)(1/sinθ)(sinθ)(1/cosθ)
= (1/2sinθcosθ)(sinθ/cosθ)
= (1/sin2θ) tanθ
= cosec2θ tanθ
(Shown)
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
sometimes if ur stuck, try solving both sides halfway like what I did to get a better idea of what to do next
Date Posted:
3 years ago
360 Tutoring Program