If sinθ+cosθ=m and secθ+cosecθ=n, then nm+1m-1=
m
n
2m
2n
Solve by substituting the values in given expression:
Put the given values sinθ+cosθ=m and secθ+cosecθ=n in the required expression, then
nm+1m-1=secθ+cosecθsinθ+cosθ+1sinθ+cosθ-1
Apply the algebraic identity a2-b2=(a+b)(a-b), then
nm+1m-1=secθ+cosecθsinθ+cosθ2-12
Now, applying the algebraic identity (a+b)2=a2+b2+2ab, we get
∴n+m+1m-1=secθ+cosecθsin2θ+cos2θ+2sinθcosθ-1=secθ+cosecθ1+2sinθcosθ-1=1cosθ+1sinθ2sinθcosθ=2sinθcosθcosθ+2sinθcosθsinθ=2sinθ+cosθ=2m
Hence, the correct option is (C).
If secθ=m and tanθ=n, then 1mm+n+1m+n is equals to
If tan(θ)+sin(θ)=m and tan(θ)-sin(θ)=n then