Question

If \[\begin{bmatrix}1 & - \tan \theta \\ \tan \theta & 1\end{bmatrix} \begin{bmatrix}1 & \tan \theta \\ - \tan \theta & 1\end{bmatrix} - 1 = \begin{bmatrix}a & - b \\ b & a\end{bmatrix}\], then _______________ .

पर्याय

• \[a = 1, b = 1\]

• \[a = \cos 2 \theta, b = \sin 2 \theta\]

• \[a = \sin 2 \theta, b = \cos 2 \theta\]

• None of these

Answer 1

\[a = \cos 2 \theta, b = \sin 2 \theta\]

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

Given:-

\[ \begin{bmatrix}1 & - \tan\theta \\ \tan\theta & 1\end{bmatrix} \begin{bmatrix}1 & \tan\theta \\ - \tan\theta & 1\end{bmatrix}^{- 1} = \begin{bmatrix}a & - b \\ b & a\end{bmatrix}\]

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

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

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

On comparing, we get

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

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

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