If 1 & - & 1 1 & - & 1 -1= a & -b b & a , then : - Vidyadip

MCQ

If $\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}-1=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix},$ then :

A
$\text{a}=1,\text{b}=1$
✓
$\text{a}=\cos2\theta,\text{b}=\sin2\theta$
C
$\text{a}=\sin2\theta,\text{b}=\cos2\theta$
D
None of these.

✓

Answer

Correct option: B.

$\text{a}=\cos2\theta,\text{b}=\sin2\theta$

$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}^{-1}=\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}$
$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{c} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} \frac{1-\tan^2\theta}{\sec^2\theta} & \frac{-2\tan\theta}{\sec^2\theta} \\ \frac{2\tan\theta}{\sec^2\theta} & \frac{1-\tan^2\theta}{\sec^2\theta} \end{bmatrix}=\begin{bmatrix}\text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$
On comparing, we get
$\text{a}=\frac{1-\tan^2\theta}{\sec^2\theta}$ and ${b}=\frac{2\tan\theta}{\sec^2\theta}$
$\Rightarrow\text{a}=\cos^2\theta-\sin^2\theta$ and ${b}=2\sin\theta\cos\theta$
$\Rightarrow\text{a}=\cos2\theta$ and ${b}=\sin2\theta$

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