If $$Cos\ \theta + Sin\ \theta$$ = m and $$Sec\ \theta + Cosec\ \theta$$= n then the value of n($$m^{2}-1$$) is equal to:

Given : $$cos\ \theta + sin\ \theta=m$$ --------------(i)

Squaring both sides, we get :

=> $$(cos\ \theta + sin\ \theta)^2=(m)^2$$

=> $$cos^2\ \theta+sin^2\ \theta+2sin\ \theta.cos\ \theta=m^2$$

=> $$1+2sin\ \theta.cos\ \theta=m^2$$

=> $$sin\ \theta.cos\ \theta=\frac{m^2-1}{2}$$ -------------(ii)

Also, it is given that : $$sec\ \theta + cosec\ \theta=n$$

=> $$\frac{1}{cos\ \theta}+\frac{1}{sin\ \theta}=n$$

=> $$\frac{sin\ \theta+cos\ \theta}{sin\ \theta.cos\ \theta}=n$$

Using equations (i) and (ii), => $$m=\frac{m^2-1}{2}\times n$$

=> $$n(m^2-1)=2m$$

=> Ans - (D)