Question

If $\left[\begin{array}{cc}1& -\tan \mathrm{\theta }\\ \tan \mathrm{\theta }& 1\end{array}\right]\left[\begin{array}{cc}1& \tan \mathrm{\theta }\\ -\tan \mathrm{\theta }& 1\end{array}\right]-1=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$, then

(a) $a=1,b=1$
(b) $a=\cos 2\mathrm{\theta },b=\sin 2\mathrm{\theta }$
(c) $a=\sin 2\mathrm{\theta },b=\cos 2\mathrm{\theta }$
(d) none of these

Answer 1

(b) $a=\cos 2\mathrm{\theta },b=\sin 2\mathrm{\theta }$

$$\begin{gathered} {\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}=\frac{1}{{\sec }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right] \\ \mathrm{Given}: \\ \left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]\frac{1}{{\sec }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right] \\ \Rightarrow \frac{1}{{\sec }^2\theta }\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}\frac{1-{\tan }^2\theta }{{\sec }^2\theta }& \frac{-2\tan \theta }{{\sec }^2\theta }\\ \frac{2\tan \theta }{{\sec }^2\theta }& \frac{1-{\tan }^2\theta }{{\sec }^2\theta }\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right] \\ \mathrm{On}\mathrm{comparing},\mathrm{we}\mathrm{get} \\ a=\frac{1-{\tan }^2\theta }{{\sec }^2\theta }\mathrm{and}b=\frac{2\tan \theta }{{\sec }^2\theta } \\ \Rightarrow a={\cos }^2\theta -{\sin }^2\theta \mathrm{and}b=2\sin \theta \cos \theta \\ \Rightarrow a=\cos 2\theta \mathrm{and}b=\sin 2\theta \end{gathered}$$