JEE Main2019Math
Q.
A value of (0, /3), for which \left| {\begin{array}{*{20}{c}} {1 + {{\cos }^2}\theta }{{{\sin }^2}\theta }{4\cos 6\theta } \\ {{{\cos }^2}\theta }{1 + {{\sin }^2}\theta }{4\cos 6\theta } \\ {{{\cos }^2}\theta }{{{\sin }^2}\theta }{1 + 4\cos 6\theta } \end{array}} \right| = \,0, is
- A
- B
- C
- D
Solution
