The correct option is A √3sinϑ−cosϑ=2
A)√3sinθ−cosθ=2⇒(√3sinθ)2=(2+cosθ)2⇒√3sinθ=2+cosθ⇒(√3sinθ)2=(2+cosθ)2⇒3sin2θ=4+cos2θ+4cosθ⇒3(1−cos2θ)=4+cos2θ+4cosθ⇒3−3cos2θ−4−cos2θ−4cosθ=0⇒−4cos2θ−4cosθ−1=0⇒cos2θ+cosθ+14=0cos=−1±√(1)2−4×1×142×1⇒cosθ=cos(π=π3)⇒cosθ=cos(2π3)⇒θ=2nπ±2π3(B)cosθ+sinθ=√2⇒1√2cosθ+1√2sinθ=1⇒sinπ4cosθ+cosπ4sinθ⇒sin(θ+π4)=1⇒θ+π4=π4+2πn⇒θ=π4+2πn(C)Hasnosolution(D)cosecθ−secθ⇒cosecθ⋅secθ⇒1sinθ−1cosθ=1sinθcosθ⇒cosθ−sinθ=1⇒(cosθ−sinθ)=1⇒cos2θ+sin2θ−2cosθsinθ=1⇒1−sin2θ=1⇒sin2θ=0⇒2θ=0+2πn⇒θ=πn