If 1 & - & 1 1 & - & 1 -1= a & -b b & a , then: 1. a=1,b=1 2. a= 2 ,b= 2 3. a= 2 ,b= 2 4. None of these.

2. a= 2 ,b= 2 Solution: 1 & - & 1 ^ -1 =1 ^2 1 & - & 1 ^ -1 1 & - & 1 1 & - & 1 ^ -1 = a & -b c & a 1 & - & 1 1 ^2 1 & - & 1 ^ -1 = a & -b b & a 1 ^2 1 & - & 1 1 & - & 1 ^ -1 = a & -b b & a 1- ^2 ^2 & -2 ^2 2 ^2 & 1- ^2 ^2 = a & -b b & a On comparing, we get a=1- ^2 ^2 and b=2 ^2 = ^2 - ^2 and b=2 = 2 and b= 2

If 1 & - & 1 1 & - & 1 -1= a & -b b & a , then: 1. a=1,b=1 2. a= …

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Question

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:

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

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Answer

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

Solution:

$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}^{-1}=\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}$

$\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{c} & \text{a} \end{bmatrix}$

$\Rightarrow\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$

$\Rightarrow\frac{1}{\sec^2\theta}\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{-1}=\begin{bmatrix} \text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$

$\Rightarrow\begin{bmatrix} \frac{1-\tan^2\theta}{\sec^2\theta} & \frac{-2\tan\theta}{\sec^2\theta} \\ \frac{2\tan\theta}{\sec^2\theta} & \frac{1-\tan^2\theta}{\sec^2\theta} \end{bmatrix}=\begin{bmatrix}\text{a} & -\text{b} \\ \text{b} & \text{a} \end{bmatrix}$

On comparing, we get

$\text{a}=\frac{1-\tan^2\theta}{\sec^2\theta}\text{ and b}=\frac{2\tan\theta}{\sec^2\theta}$

$\Rightarrow\text{a}=\cos^2\theta-\sin^2\theta\text{ and b}=2\sin\theta\cos\theta$

$\Rightarrow\text{a}=\cos2\theta\text{ and b}=\sin2\theta$