If $$1$$−$$tan$$⁡θ$$tan$$⁡θ$$11tan$$⁡θ−$$tan$$⁡θ$$1$$−$$1=a$$−bba then

Correct option is 2a=cos2θ, b=sin2θ

Questions

If $(\begin{array}{cc} 1 & - \tan θ \\ \tan θ & 1 \end{array}) {(\begin{array}{cc} 1 & \tan θ \\ - \tan θ & 1 \end{array})}^{- 1} = (\begin{array}{cc} a & - b \\ b & a \end{array})$ then

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a=b=1

a=cos2θ, b=sin2θ

a=sin2θ, b=cos2θ

a=1, b=sin2θ

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detailed solution

Correct option is B

a−bba=1−tan⁡θtan⁡θ1cos2⁡θ−sin⁡θcos⁡θsin⁡θcos⁡θcos2⁡θ =cos⁡2θ−sin⁡2θsin⁡2θcos⁡2θ ⇒a=cos⁡2θ,b=sin⁡2θ

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If 1−tan⁡θtan⁡θ11tan⁡θ−tan⁡θ1−1=a−bba then

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If $(\begin{array}{cc} 1 & - \tan θ \\ \tan θ & 1 \end{array}) {(\begin{array}{cc} 1 & \tan θ \\ - \tan θ & 1 \end{array})}^{- 1} = (\begin{array}{cc} a & - b \\ b & a \end{array})$ then