If `[(1,-tan theta),(tan theta,1)][(1,tan theta),(-tan theta,1)]^(-1)=[(a,-b),(b,a)],` then

Options

A. `a= cos 2 theta`
B. `a=1`
C. `b= sin 2 theta`
D. `b=-1`

Explanation

We have,
`[(1,tan theta),(-tan theta,1)]^(-1) =1/(1+ tan^(2) theta) [(1,- tan theta),(tan theta,1)]`
`implies [(a,-b),(b,a)]=cos^(2) theta [(1,-tan theta),(tan theta,1)][(1,-tan theta),(tan theta,1)]`
`=cos^(2) theta [(1-tan^(2) theta,-2 tan theta),(2 tan theta,1-tan^(2) theta)]`
`=[(cos 2 theta,- sin 2 theta),(sin 2 theta,cos 2 theta)]`
`:. a=cos 2 theta, b= sin 2 theta`