If sinθ+cosθ=2cosθ then cosθ−sinθis equal to
2 cos θ
2 sin θ
2 (cos θ + sin θ)
none of these
We have,
sinθ+cosθ=2cosθ⇒1+sin2θ=2cos2θ⇒1−sin2θ=2−2cos2θ⇒(cosθ−sinθ)2=2sin2θ⇒cosθ−sinθ=2sinθ