Q:

What is the condition for a body on an inclined plane(θ degree with horizontal) to fall freely?

a body is lying on the frictionless wedge as shown in the figure. The necessary horizontal acceleration "a" that must be given on the wedge such that the body  falls freely is acotθ . How to get this?
Is the following approach correct?
mgcosθ-N=masinθ ⇒ mgcosθ-masinθ=N ⇒ mgcosθ=masinθ [since N=0 during free fall] therefore a=gcotθ

A:

Your answer is correct.  Why do you think it means the object does not fall straight down?  You found a condition such that N=0, and that's the only influence of the plane on the block.  So if N=0, the situation has to be identical to having no plane at all, which of course will lead to motion straight downward if the initial velocity is zero.  Note you are solving for a, not v, so the initial v can be whatever you want, including v=0.

