If cosec θ – sin θ = b, sec θ – cos θ = a sinθcos θ = ???

I'm going to use x instead of
theta.

It is given that cosec x - sin x = b and sec x - cos
x = a.

We need to find sin x *
cos

sec x = 1/cos x and cosec x  = 1/ sin
x

cosec x - sin x = b and sec x - cos x =
a

=> 1/ sin x - sin x = b and 1/cos x - cos x =
a

b = [1 - (sin x)^2]/sin x and a = [1 - (cos x)^2]/cos
x

=> b = (cos x)^2 / sin x and a = (sin x)^2/ cos
x

b *a = [(cos x)^2 / sin x]*[(sin x)^2/ cos
x]

=> [(cos x)^2 *(sin x)^2 / sin x*cos
x]

=> sin x * cos
x

This gives us (sin x)*(cos x) =
a*b