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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Matrices
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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 find $a,b$

(A) $\;a=\cos \theta,b=\sin \theta$ (B) $\;a=\cos 2\theta,b=\sin 2\theta$ (C) $\;a=\tan\theta,b=\cot\theta$ (D) $\;None\;of\;these$
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1 Answer

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$\begin{bmatrix}1 &\tan\theta\\-\tan\theta&1\end{bmatrix}^{-1}=\large\frac{1}{1+\tan^2\theta}$$\begin{bmatrix}1 &-\tan\theta\\\tan\theta&1\end{bmatrix}$
$\begin{bmatrix}a &-b\\b &a\end{bmatrix}=\cos^2\theta\begin{bmatrix}1&-\tan\theta\\\tan\theta&1\end{bmatrix}\begin{bmatrix}1 &-\tan\theta\\\tan\theta&1\end{bmatrix}$
$\qquad\qquad=\cos^2\theta\begin{bmatrix}1-\tan^2\theta&-2\tan\theta\\2\tan\theta&1-\tan^2\theta\end{bmatrix}$
$\qquad\qquad=\begin{bmatrix}\cos 2\theta &-\sin 2\theta\\\sin 2\theta &\cos 2\theta\end{bmatrix}$
$\Rightarrow a=\cos 2\theta$,$b=\sin 2\theta$
Hence (b) is the correct answer.
answered Nov 19, 2013 by sreemathi.v
 

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