If $A = [\begin{array}{cc} 1 & \tan x \\ - \tan x & 1 \end{array}]$ , then A^TA^-1 is

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

none of these

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

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

answer is B.

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Detailed Solution

$| A | = |\begin{array}{cc} 1 & \tan x \\ - \tan x & 1 \end{array}| = 1 + \tan^{2} x \neq 0$

So A is invertible. Also,

$adj A = {[\begin{array}{cc} 1 & \tan x \\ - \tan x & 1 \end{array}]}^{T} = [\begin{array}{cc} 1 & - \tan x \\ \tan x & 1 \end{array}]$

Now, $A^{- 1} = \frac{1}{| A |} adj A$

$\begin{array}{l} = \frac{1}{(1 + \tan^{2} x)} [\begin{array}{cc} 1 & - \tan x \\ \tan x & 1 \end{array}] \\ = [\begin{array}{cc} \frac{1}{1 + \tan^{2} x} & \frac{- \tan x}{1 + \tan^{2} x} \\ \frac{\tan x}{1 + \tan^{2} x} & \frac{1}{1 + \tan^{2} x} \end{array}] \end{array}$

$\begin{matrix} ∴ A^{T} A^{- 1} = [\begin{array}{cc} 1 & - \tan x \\ \tan x & 1 \end{array}] [\begin{array}{cc} \frac{1}{1 + \tan^{2} x} & \frac{- \tan x}{1 + \tan^{2} x} \\ \frac{\tan x}{1 + \tan^{2} x} & \frac{1}{1 + \tan^{2} x} \end{array}] \\ = [\begin{array}{cc} \frac{1 - \tan^{2} x}{1 + \tan^{2} x} & \frac{- 2 \tan x}{1 + \tan^{2} x} \\ \frac{2 \tan x}{1 + \tan^{2} x} & \frac{1 - \tan^{2} x}{1 + \tan^{2} x} \end{array}] \\ = [\begin{array}{cc} \cos 2 x & - \sin 2 x \\ \sin 2 x & \cos 2 x \end{array}] \end{matrix}$