Mathematics · 2019
JEE Main · 12 April 2019, Shift 2 · Q64
A value of θ ∈(0, π / 3), for which |1+ cos^2 θ, sin^2 θ, 4 cos 6 θ; cos^2 θ, 1+ sin^2 θ, 4 cos 6 θ; cos^2 θ, sin^2 θ, 1+4 cos 6 θ| =0, is:
A value of $\displaystyle \theta \in(0, \pi / 3)$, for which
$$\left|\begin{array}{ccc}
1+\cos ^2 \theta & \sin ^2 \theta & 4 \cos 6 \theta \\
\cos ^2 \theta & 1+\sin ^2 \theta & 4 \cos 6 \theta \\
\cos ^2 \theta & \sin ^2 \theta & 1+4 \cos 6 \theta
\end{array}\right|=0, \text { is : }
$$
Official answer
(2)
$\displaystyle \frac{\pi}{9}$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.