If A=[sinθ−cosθcosθsinθ] and B=[cosθsinθ−sinθcosθ] then (sinθ)A+(cosθ)B=

(sinθ)A+(cosθ)B=[1001] ⇒[sin2θ−sinθcosθsinθcosθsin2θ]+[cos2θsinθcosθ−sinθcosθcos2θ] =[1001]

If A=[sinθ−cosθcosθsinθ] and B=[cosθsinθ−sinθcosθ] then (sinθ)A+(cosθ)B=

(sinθ)A+(cosθ)B=[1001] ⇒[sin2θ−sinθcosθsinθcosθsin2θ]+[cos2θsinθcosθ−sinθcosθcos2θ] =[1001]

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If $A = [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$ and $B = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ then $(\sin θ) A + (\cos θ) B =$

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[\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}]

[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

[\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}]

[\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]

answer is B.

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(\sin θ) A + (\cos θ) B = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

$\Rightarrow [\begin{matrix} \sin^{2} θ & - \sin θ \cos θ \\ \sin θ \cos θ & \sin^{2} θ \end{matrix}] + [\begin{matrix} \cos^{2} θ & \sin θ \cos θ \\ - \sin θ \cos θ & \cos^{2} θ \end{matrix}]$

$= [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

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If A=[sinθ−cosθcosθsinθ] and B=[cosθsinθ−sinθcosθ] then (sinθ)A+(cosθ)B=

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If $A = [\begin{matrix} \sin θ & - \cos θ \\ \cos θ & \sin θ \end{matrix}]$ and $B = [\begin{matrix} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{matrix}]$ then $(\sin θ) A + (\cos θ) B =$