If A=\left[\begin{array}{cc}1 & \tan x \ -\tan x & 1]\end{array}, then A^T \cdot A^{-1}=

Correct Answer is : [cos2x−sin2xsin2xcos2x]

\left[\begin{array}{cc}-\cos 2 x & \sin \\\2 x \ - \sin \\\2 x & \cos 2 x]\end{array}

\left[\begin{array}{cc}\cos 2 x & - \sin \\\2 x \ \sin \\\2 x & \cos 2 x]\end{array}

\left[\begin{array}{cc}\cos 2 x & \sin \\\2 x \ - \sin \\\2 x & \cos 2 x]\end{array}

\left[\begin{array}{cc}\cos 2 x & - \sin \\\2 x \ - \sin \\\2 x & \cos 2 x]\end{array}

MHTCET 13 May 2023 Shift 2 Solved Paper

Section: Mathematics

Question : 61 of 150

Marks: +1, -0

If A=[

1	tan‌x
−tan‌x	1

], then AT⋅A−1=

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−cos‌2‌x	sin‌2x
−sin‌2x	cos‌2‌x

cos‌2‌x	−sin‌2x
sin‌2x	cos‌2‌x

cos‌2‌x	sin‌2x
−sin‌2x	cos‌2‌x

cos‌2‌x	−sin‌2x
−sin‌2x	cos‌2‌x