Find the range of `f(x)=1/((cosx-3)^2+(sinx+4)^2)`

Explanation

`f(x)=(1)/((cos x-3)^(2)+(sinx+4)^(2))`
`=(1)/(cos^(2)-6cosx+9+sin^(2)x+8sinx+16)`
`(1)/(8sinx-6cosx+26)`
Now `8sinx-6cos x in[-sqrt((8)^(2)+(-6)^(2)),sqrt((8)^(2)+(-6)^(2))]`
`therefore 8 sin x -6 cos x in [ -10,10]`
`therefore 8 sin x -6 cos x+26in[16,36]`
`therefore (1)/(36)le(1)/((cosx-3)^(3)+(sin x+4)^(2))le(1)/(16)`
thus range of `f(x)` is `[(1)/(36),(1)/(16)]`