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If ${\Delta _1} = \left| {\begin{array}{*{20}{c}}
x&{\sin \,\theta }&{\cos \,\theta } \\
{\sin \,\theta }&{ - x}&1 \\
{\cos \,\theta }&1&x
\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}}
x&{\sin \,2\theta }&{\cos \,\,2\theta } \\
{\sin \,2\theta }&{ - x}&1 \\
{\cos \,\,2\theta }&1&x
\end{array}} \right|$, $x \ne 0$ ; then for all $\theta \in \left( {0,\frac{\pi }{2}} \right)$
${\Delta _1} - {\Delta _2} = - 2{x^3}$
${\Delta _1} + {\Delta _2} = - 2({x^3} + x - 1)$
${\Delta _1} - {\Delta _2} = x\left( {\cos \,2\theta - \cos \,4\theta } \right)$
${\Delta _1} + {\Delta _2} = - 2{x^3}$
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}$
${\Delta _1} + {\Delta _2} = – 2{x^3}$
