Q95 / 180 · COMEDK 2018 · Mathematics

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

  • A
    $a=1=b$
  • Correct answer
    B
    $a=\cos 2 \theta, b=\sin 2 \theta$
  • C
    $a=\sin 2 \theta, b=\cos 2 \theta$
  • D
    $a=\cos \theta, b=\sin \theta$

Solution

We have, $\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\frac{1}{1+\tan ^{2} \theta} \quad\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$ $\left[\because\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\right]$ Now, $\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left(\frac{1}{1+\tan ^{2} \theta}\right)$ $\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow \frac{1}{1+\tan ^{2} \theta}\left[\begin{array}{cc}1-\tan ^{2} \theta & -2 \tan \theta \\ 2 \tan \theta & 1-\tan ^{2} \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$ $\therefore \quad a=\cos 2 \theta, b=\sin 2 \theta$

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

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