The value of $$\frac{{2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)}}{{{{\cos }^4}\theta - {{\sin }^4}\theta - 2{{\cos }^2}\theta }}{\text{ is:}}$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{{2\left( {{{\sin }^6}\theta + {{\cos }^6}\theta } \right) - 3\left( {{{\sin }^4}\theta + {{\cos }^4}\theta } \right)}}{{{{\cos }^4}\theta - {{\sin }^4}\theta - 2{{\cos }^2}\theta }} \cr & = \frac{{2\left( {1 - 3{{\sin }^2}\theta .{{\cos }^2}\theta } \right) - 3\left( {1 - 2{{\sin }^2}\theta .{{\cos }^2}\theta } \right)}}{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) - 2{{\cos }^2}\theta }} \cr & = \frac{{2 - 6{{\sin }^2}\theta .{{\cos }^2}\theta - 3 + 6{{\sin }^2}\theta .{{\cos }^2}\theta }}{{ - \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}} \cr & = \frac{{ - 1}}{{ - 1}} \cr & = 1 \cr} $$