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यदि $\Delta_{1}=\left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & - x & 1 \\ \cos \theta & 1 & x \end{array}\right|$ तथा $\Delta_{2}=\left|\begin{array}{ccc}x & \sin 2 \theta & \cos 2 \theta \\ -\sin 2 \theta & -x & 1 \\ \cos 2 \theta & 1 & x\end{array}\right|, x \neq 0$; तो सभी $\theta \in\left(0, \frac{\pi}{2}\right)$ के लिए
${\Delta _1} - {\Delta _2} = - 2{x^3}$
${\Delta _1} + {\Delta _2} = - 2({x^3} + x - 1)$
${\Delta _1} - {\Delta _2} = x\left( {\cos \,2\theta - \cos \,4\theta } \right)$
${\Delta _1} + {\Delta _2} = - 2{x^3}$
Solution
${\Delta _1} = \left| {\begin{array}{*{20}{c}}
x&{\sin \theta }&{\cos \theta }\\
{ – \sin \theta }&{ – x}&1\\
{\cos \theta }&1&x
\end{array}} \right|$
$ = x\left( { – {x^2} – 1} \right) – \sin \theta \left( { – x\sin \theta – \cos \theta } \right) + \cos \theta \left( { – \sin \theta + x\cos \theta } \right)$
$ \Rightarrow – {x^3}$
${\Delta _2} = \left| {\begin{array}{*{20}{c}}
x&{\sin 2\theta }&{\cos 2\theta }\\
{ – \sin 2\theta }&{ – x}&1\\
{\cos 2\theta }&1&x
\end{array}} \right|$
$ \Rightarrow – {x^3}$
${\Delta _1} + {\Delta _2} = – 2{x^3}$
