If 1-tan θtan θ1 1tan θ-tan θ1=a-bba, then

a-bba=1-tanθtanθ11tanθ-tanθ1= 1+tan2θtanθ-tanθtanθ-tanθ1+tan2θ ⇒a=sec2θ, b=0

If 1-tan θtan θ1 1tan θ-tan θ1=a-bba, then

a-bba=1-tanθtanθ11tanθ-tanθ1= 1+tan2θtanθ-tanθtanθ-tanθ1+tan2θ ⇒a=sec2θ, b=0

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If $[\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] [\begin{matrix} 1 & \tan θ \\ - \tan θ & 1 \end{matrix}] = [\begin{matrix} a & - b \\ b & a \end{matrix}]$ , then

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a=1,b=1

$a = cos2 θ, b = sin2 θ$

$a = sin2 θ, b= cos2 θ$

$a = \sec^{2} θ, b = 0$

answer is D.

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Detailed Solution

$[\begin{matrix} a & - b \\ b & a \end{matrix}] = [\begin{matrix} 1 & - \tan θ \\ \tan θ & 1 \end{matrix}] [\begin{matrix} 1 & \tan θ \\ - \tan θ & 1 \end{matrix}] = [\begin{matrix} 1 + \tan^{2} θ & \tan θ - \tan θ \\ \tan θ - \tan θ & 1 + \tan^{2} θ \end{matrix}] \Rightarrow a = \sec^{2} θ, b = 0$