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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
MCQ
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
    $a=1=b$
  • B
    $a=\cos 2 \theta, b=\sin 2 \theta$
  • C
    $a=\sin 2 \theta, b=\cos \theta$
  • D
    $a=\cos \theta, b=\sin \theta$
✓

Answer

We have,
$
\begin{array}{l}
{\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]} \\
\text { Now, }\left[\begin{array}{cc}
1 & \tan \theta \\
-\tan \theta & 1
\end{array}\right]^{-1}=\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc}
1 & -\tan \theta \\
\tan \theta & 1
\end{array}\right] \\
\quad=\left[\begin{array}{cc}
\cos ^2 \theta & -\tan \theta \cos ^2 \theta \\
\cos ^2 \theta \tan \theta & \cos ^2 \theta
\end{array}\right] \\
\Rightarrow\left[\begin{array}{cc}
1 & -\tan \theta \\
\tan \theta & 1
\end{array}\right]\left[\begin{array}{cc}
\cos ^2 \theta & -\tan \theta \cos ^2 \theta \\
\cos ^2 \theta \tan \theta & \cos ^2 \theta
\end{array}\right]=\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right] \\
\Rightarrow\left[\begin{array}{cc}
\cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta & -2 \tan \theta \cos ^2 \theta \\
2 \tan \theta \cos ^2 \theta & \cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta
\end{array}\right] \\
\quad=\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right] \\
\therefore a=\cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta \text { and } b=2 \tan ^2 \cos ^2 \theta \\
\Rightarrow a=\cos ^2 \theta\left(1-\frac{\sin ^2 \theta}{\cos ^2 \theta}\right) \text { and } b=\frac{2 \sin \theta}{\cos \theta} \cdot \cos ^2 \theta \\
\Rightarrow a=\cos ^2 \theta-\sin ^2 \theta=\cos 2 \theta \text { and } b=2 \sin \theta \cos \theta=\sin 2 \theta
\end{array}
$

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