Answer:
2sin²θ - tan²θ
Step-by-step explanation:
Given
tan²θcos(2θ)
Required
Simplify
We start by simplifying cos(2θ)
cos(2θ) = cos(θ+θ)
From Cosine formula
cos(A+A) = cosAcosA - sinAsinA
cos(A+A) = cos²A - sin²A
By comparison
cos(2θ) = cos(θ+θ)
cos(2θ) = cos²θ - sin²θ ----- equation 1
Recall that cos²θ + sin²θ = 1
Make sin²θ the subject of formula
sin²θ = 1 - cos²θ
Substitute sin²θ = 1 - cos²θ in equation 1
cos(2θ) = cos²θ - (1 - cos²θ)
cos(2θ) = cos²θ - 1 +cos²θ
cos(2θ) = cos²θ + cos²θ - 1
cos(2θ) = 2cos²θ - 1
Substitute 2cos²θ - 1 for cos(2θ) in the given question
tan²θcos(2θ) becomes
tan²θ(2cos²θ - 1)
Open brackets
2cos²θtan²θ - tan²θ
------------------------
Simplify tan²θ
tan²θ = (tanθ)²
Recall that tanθ = sinθ/cosθ
So, we have
tan²θ = (sinθ/cosθ)²
tan²θ = sin²θ/cos²θ
------------------------
Substitute sin²θ/cos²θ for tan²θ
2cos²θtan²θ - tan²θ becomes
2cos²θ(sin²θ/cos²θ) - tan²θ
Open bracket (cos²θ will cancel out cos²θ) to give
2(sin²θ) - tan²θ
2sin²θ - tan²θ
Hence, the simplification of tan²θcos(2θ) is 2sin²θ - tan²θ
Option E is correct
= 18 unit²