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If cosθ+sinθ=mcos heta+sin heta=m and secθ+cosecθ=nsec heta+operatorname{cosec} heta=n, then what is the value of n2(m21)?frac{n}{2}left(m^{2}-1 ight) ?

a

m

b

2n

c

2m

d

mn

Answer : Option A
Explanation :

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)

=12(1cosθ+1sinθ)=sinθ+cosθ2cosθsinθ=frac{1}{2}left(frac{1}{cos heta}+frac{1}{sin heta} ight)=frac{sin heta+cos heta}{2 cos heta cdot sin heta}

n2(m21)=sinθ+cosθ2cosθsinθ×2cosθsinθ 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

=sinθ+cosθ=m.=sin heta+cos heta=m .

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