The value of $\theta $ lying between $\theta = 0$ and $\frac{\pi }{2}$ and satisfying the equation $\left| {\begin{array}{*{20}{c}} {1 + {{\sin }^2}{\rm{\theta }}}&{{{\cos }^2}{\rm{\theta }}}&{4\sin 4{\rm{\theta }}}\\ {{{\sin }^2}{\rm{\theta }}}&{1 + {{\cos }^2}{\rm{\theta }}}&{4\sin 4{\rm{\theta }}}\\ {{{\sin }^2}{\rm{\theta }}}&{{{\cos }^2}{\rm{\theta }}}&{1 + 4\sin 4{\rm{\theta }}} \end{array}} \right|$ is

JEE MATHEMATICS MCQ: MATRICES

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Application +4 | -1

The value of

\theta

lying between

\theta = 0

and

\frac{\pi }{2}

and satisfying the equation

\left| {\begin{array}{cccccccccccccccccccc} {1 + {{\sin }^2}\mathrm{\theta }}&{{{\cos }^2}\mathrm{\theta }}&{4\sin 4\mathrm{\theta }}\\ {{{\sin }^2}\mathrm{\theta }}&{1 + {{\cos }^2}\mathrm{\theta }}&{4\sin 4\mathrm{\theta }}\\ {{{\sin }^2}\mathrm{\theta }}&{{{\cos }^2}\mathrm{\theta }}&{1 + 4\sin 4\mathrm{\theta }} \end{array}} \right|

View answer and explanation

Correct option: A

\frac{{7\pi }}{{24}}

Option A (

\frac{{7\pi}}{{24}}

) is correct because it satisfies the given determinant equation within the specified range of

\theta

after simplification and solving for

\theta