- Home
- Standard 12
- Mathematics
$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to
$1/2$
$1$
$-1/2$
$-1$
Solution
(c) $\left| {\,\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$
Using ${C_1} \to {C_1} – {C_2},{C_2} \to {C_2} – {C_3}$
==> $\left| {\,\begin{array}{*{20}{c}}1&0&{{{\sin }^2}\theta }\\{ – 1}&1&{{{\cos }^2}\theta }\\0&{ – 1}&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$
==> $2\,(1 + 2\sin 4\theta ) = 0 \Rightarrow \sin 4\theta = \frac{{ – 1}}{2}$.
