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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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If $A(\theta)=\begin{bmatrix}1&\tan\theta\\-\tan\theta&1\end{bmatrix}$ and $AB=I$ then $(\sec^2\theta)B$ is equal to

$(a)\;A(\theta)\qquad(b)\;A(\large\frac{\theta}{2})$$\qquad(c)\;A(-\theta)\qquad(d)\;A(-\large\frac{\theta}{2})$

Can you answer this question?
 
 

1 Answer

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$A(\theta)=\begin{bmatrix}1&\tan \theta\\-\tan\theta&1\end{bmatrix}$
$AB=I$
$\Rightarrow B=A^{-1}$
$\Rightarrow \large\frac{1}{1+\tan^2\theta}$$\begin{bmatrix}1&-\tan\theta\\\tan\theta&1\end{bmatrix}$
$1+\tan^2\theta=\sec^2\theta$
$\Rightarrow \large\frac{1}{\sec^2\theta}$$\begin{bmatrix}1&-\tan\theta\\\tan\theta&1\end{bmatrix}$
$\Rightarrow (\sec^2\theta)B=\begin{bmatrix}1&-\tan\theta\\\tan\theta&1\end{bmatrix}$
$\qquad\qquad\;\;= A(-\theta)$
Hence (c) is the correct answer.
answered Nov 22, 2013 by sreemathi.v
 

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