Q56 / 100 · BITSAT 2022 · Mathematics

$\text { If }\left[\begin{array}{cc} 1 & -\tan \theta \\ \tan \theta & 1 \end{array}\right]\left[\begin{array}{cc} 1 & \tan \theta \\ -\tan \theta & 1 \end{array}\right]^{-1}=\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right],$ then

  • A
    $a=1, b=1$
  • B
    $a=\sin 2 \theta, b=\cos 2 \theta$
  • Correct answer
    C
    $a=\cos 2 \theta, b=\sin 2 \theta$
  • D
    None of these

Solution

Given, $\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right] \cdot \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$ $=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1-\tan ^2 \theta & -2 \tan \theta \\ 2 \tan \theta & 1-\tan ^2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow \quad\left[\begin{array}{cc}\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} & \frac{2 \tan \theta}{1+\tan ^2 \theta} \\ \frac{2 \tan \theta}{1+\tan ^2 \theta} & \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow \quad a=\cos 2 \theta, b=\sin 2 \theta$

Hence, the correct answer is option: 'C'.

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