If \(U =\) cosθ00cosθ, \(V =\) sinθ00sinθ then deduce that \(U^2 + V^2 = I\).
 
Answer:
Answer variants:
\(2 cos \theta\)
cos2θ
\(1\)
\(0\)
\(2 sin \theta\)
sin2θ
\(U^2 =\) iiii
 
\(V^2 =\) iiii
 
\(U^2 + V^2\) \(=\) iiii [By the trigonometric identity ]
 
\(U^2 + V^2 = I\)
 
Hence, proved.