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 }{4sin 4theta }&{4sin 4theta }&{1 + 4sin 4theta

Q: $\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ તો $\sin\, 4\theta $ મેળવો.

$\left| {\,\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}\,} \right| = 0$ Using ${C_1} \to {C_1} - {C_2},{C_2} \to {C_2} - {C_3}$ $ \Rightarrow \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$ $ \Rightarrow 2\,(1 + 2\sin 4\theta ) = 0$ $ \Rightarrow \sin 4\theta = \frac{{ - 1}}{2}$.

MCQ

ગુજરાતી માધ્યમ

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ તો $\sin\, 4\theta $ મેળવો.

A$1/2$
B$1$
C$-1/2$
D$-1$

Difficult

ધોરણ 12 સાયન્સ
ગણિત
3 and 4 . determinant and metrices
JEE

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