JEE MAIN - Mathematics Hindi (2019 - 10th April Morning Slot - No. 2)
यदि $${\Delta _1} = \left| {\matrix{
x & {\sin \theta } & {\cos \theta } \cr
{ - \sin \theta } & { - x} & 1 \cr
{\cos \theta } & 1 & x \cr
} } \right|$$ और
$${\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|$$, $$x \ne 0$$ ;
तो सभी $$\theta \in \left( {0,{\pi \over 2}} \right)$$ के लिए :
$${\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|$$, $$x \ne 0$$ ;
तो सभी $$\theta \in \left( {0,{\pi \over 2}} \right)$$ के लिए :
$${\Delta _1} - {\Delta _2}$$ = x (cos 2$$\theta $$ – cos 4$$\theta $$)
$${\Delta _1} + {\Delta _2}$$ = - 2x3
$${\Delta _1} + {\Delta _2}$$ = – 2(x3 + x –1)
$${\Delta _1} - {\Delta _2}$$ = - 2x3
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