- Home
- Standard 12
- Mathematics
$\theta \in(0, \pi / 3)$ का एक मान, जिसके लिये $\left|\begin{array}{ccc}1+\cos ^{2} \theta & \sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & 1+\sin ^{2} \theta & 4 \cos 6 \theta \\ \cos ^{2} \theta & \sin ^{2} \theta & 1+4 \cos 6 \theta\end{array}\right|=0$ है
$\frac{\pi }{18}$
$\frac{\pi }{9}$
$\frac{7\pi }{36}$
$\frac{7\pi }{24}$
Solution
$\theta \in \left( {0,\frac{\pi }{3}} \right)$
$\left| {\begin{array}{*{20}{c}}
{1 + {{\cos }^2}\theta }&{{{\sin }^2}\theta }&{4\cos 6\theta }\\
{{{\cos }^2}\theta }&{1 + {{\sin }^2}\theta }&{4\cos 6\theta }\\
{{{\cos }^2}\theta }&{{{\sin }^2}\theta }&{1 + 4\cos 6\theta }
\end{array}} \right| = 0$
${R_2} \to {R_2} – {R_1},{R_3} \to {R_3} – {R_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
{1 + {{\cos }^2}\theta }&{{{\sin }^2}\theta }&{4\cos 6\theta }\\
{ – 1}&1&0\\
{ – 1}&0&1
\end{array}} \right| = 0$
${C_1} \to {C_1} + {C_2}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
2&{{{\sin }^2}\theta }&{4\cos 6\theta }\\
0&1&0\\
{ – 1}&0&1
\end{array}} \right| = 0$
expanding along first column
$ \Rightarrow 2\left[ {1 – 0} \right] – 1\left[ { – 4\cos 6\theta } \right] = 0$
$ \Rightarrow 2 + 4\cos 6\theta = 0$
$ \Rightarrow \cos 6\theta = – \frac{1}{2}$
$ \Rightarrow 6\theta = \frac{{2\pi }}{3}$
$ \Rightarrow \theta = \frac{\pi }{9}$