Find the value of sec2θ - \(\frac{sin^2θ-2sin^4θ}{2cos^2θ-cos^2θ}\) is?
Answer & explanation
Correct answer: option 1
⇒ sec2θ - \(\frac{sin^2θ(1-2sin^2θ)}{cos^2θ(2cos^2θ-1)}\)
We know
cos2θ - sin2θ = 2cos2θ - 1 = 1 - 2sin2θ
⇒ sec2θ - tan2θ = 1
Find the value of sec2θ - \(\frac{sin^2θ-2sin^4θ}{2cos^2θ-cos^2θ}\) is?
Correct answer: option 1
⇒ sec2θ - \(\frac{sin^2θ(1-2sin^2θ)}{cos^2θ(2cos^2θ-1)}\)
We know
cos2θ - sin2θ = 2cos2θ - 1 = 1 - 2sin2θ
⇒ sec2θ - tan2θ = 1