KeerthanaPosted on If cosθ+sinθ=mcos heta+sin heta=mcosθ+sinθ=m and secθ+cosecθ=nsec heta+operatorname{cosec} heta=nsecθ+cosecθ=n, then what is the value of n2(m2−1)?frac{n}{2}left(m^{2}-1 ight) ?2n(m2−1)? am b2n c2m dmn Answer : Option AExplanation : cosθ + sinθ = m ∴ m² – 1 = (cosθ + sinθ)² – 1 = cos²θ + sin²θ + 2cosθ. sinθ – 1 = 2 cosθ. sinθ Again, n2=12(secθ+cosecθ)frac{n}{2}=frac{1}{2}(sec heta+operatorname{cosec} heta)2n=21(secθ+cosecθ) =12(1cosθ+1sinθ)=sinθ+cosθ2cosθ⋅sinθ=frac{1}{2}left(frac{1}{cos heta}+frac{1}{sin heta} ight)=frac{sin heta+cos heta}{2 cos heta cdot sin heta}=21(cosθ1+sinθ1)=2cosθ⋅sinθsinθ+cosθ ∴n2(m2−1)=sinθ+cosθ2cosθ⋅sinθ×2cosθ⋅sinθ herefore frac{n}{2}left(m^{2}-1 ight)=frac{sin heta+cos heta}{2 cos heta cdot sin heta} imes 2 cos heta cdot sin heta∴2n(m2−1)=2cosθ⋅sinθsinθ+cosθ×2cosθ⋅sinθ =sinθ+cosθ=m.=sin heta+cos heta=m .=sinθ+cosθ=m. Rate This:NaN / 5 - 1 votesAdd comment