if Δ$$1=xsin$$θ$$cos$$θ−$$sin$$θ−$$x1cos$$θ$$1x$$ and Δ$$2=xsin2$$θ$$cos2$$θ−$$sin2$$θ−$$x1cos2$$θ$$1x$$ x≠$$0$$, then for all θ∈$$0$$,π$$2$$

Question

if Δ$$1=xsin$$θ$$cos$$θ−$$sin$$θ−$$x1cos$$θ$$1x$$ and Δ$$2=xsin2$$θ$$cos2$$θ−$$sin2$$θ−$$x1cos2$$θ$$1x$$  x≠$$0$$, then for all θ∈$$0$$,π$$2$$

Infinity Learn · Accepted Answer

We have Δ$$1=xsin$$θ$$cos$$θ$$-$$$$sin$$θ$$-x1cos$$θ$$1x$$                       $$=x-x2-1+$$$$sin$$θxsinθ$$-$$$$cos$$θ$$+$$$$cos$$θ$$sin$$θ$$+xcos$$θ                        $$=-x3$$           and  Δ$$2=xsin2$$θ$$cos2$$θ$$-sin2$$θ$$-x1cos2$$θ$$1x$$                         $$=x-x2-1+sin2$$θ$$xsin2$$θ$$-cos2$$θ$$+cos2$$θ$$sin2$$θ$$+xcos2$$θ                         $$=-x3$$ ∴ Δ$$1+$$ Δ$$2$$ $$=-2x3$$

if $Δ_{1} = |\begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix}|$ and $Δ_{2} = |\begin{matrix} x & \sin 2 θ & \cos 2 θ \\ - \sin 2 θ & - x & 1 \\ \cos 2 θ & 1 & x \end{matrix}| (x \neq 0)$ , then for all $θ \in (0, \frac{π}{2})$

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if Δ1=xsinθcosθ−sinθ−x1cosθ1x and Δ2=xsin2θcos2θ−sin2θ−x1cos2θ1x x≠0, then for all θ∈0,π2

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if $Δ_{1} = |\begin{matrix} x & \sin θ & \cos θ \\ - \sin θ & - x & 1 \\ \cos θ & 1 & x \end{matrix}|$ and $Δ_{2} = |\begin{matrix} x & \sin 2 θ & \cos 2 θ \\ - \sin 2 θ & - x & 1 \\ \cos 2 θ & 1 & x \end{matrix}| (x \neq 0)$ , then for all $θ \in (0, \frac{π}{2})$