Youif sintheta+costheta=m and sec theta+cosec theta=n , then n(m+1)(m-1) =

The given equations are sinθ + cosθ = m and secθ + cosecθ = n.

We know that sec^2θ = 1 + tan^2θ and cosec^2θ = 1 + cot^2θ.

So, secθ + cosecθ = √(1 + tan^2θ) + √(1 + cot^2θ).

We also know that tanθ = sinθ/cosθ and cotθ = cosθ/sinθ.

So, secθ + cosecθ = √(1 + (sin^2θ/cos^2θ)) + √(1 + (cos^2θ/sin^2θ)).

This simplifies to √((sin^2θ + cos^2θ)/cos^2θ) + √((sin^2θ + cos^2θ)/sin^2θ).

Since sin^2θ + cos^2θ = 1, this further simplifies to √(1/cos^2θ) + √(1/sin^2θ).

This is equal to secθ + cosecθ = n.

Now, we need to find the value of n(m^2 - 1).

Substituting the values of m and n, we get n(m^2 - 1) = (secθ + cosecθ)(sin^2θ + cos^2θ - 1).

Since sin^2θ + cos^2θ = 1, this simplifies to n(m^2 - 1) = 0.

So, n(m^2 - 1) = 0 is the solution to the given problem.