3. Show that AB - BA in each of the following cases:A-| sin θ-sin θ]

1.	3. Show that AB - BA in each of the following cases:A-\| sin θ-sin θ]B1cos φsin φ-sin φcos φand
Answer» Here AB = [cos∅cos¶ -sin∅sin¶ -sin¶cos∅+cos∅cos¶[sin∅cos¶ + cos∅sin¶. -sin∅sin¶ +cos∅cos¶ and BA = cos∅cos¶ -sin∅sin¶ -sin¶cos∅+cos∅cos¶[sin∅cos¶ + cos∅sin¶. -sin∅sin¶ +cos∅cos¶ hence AB=BA

3. Show that AB - BA in each of the following cases:A-| sin θ-sin θ]B1cos φsin φ-sin φcos φand

Answer»

Here AB = [cos∅cos¶ -sin∅sin¶ -sin¶cos∅+cos∅cos¶[sin∅cos¶ + cos∅sin¶. -sin∅sin¶ +cos∅cos¶

and BA = cos∅cos¶ -sin∅sin¶ -sin¶cos∅+cos∅cos¶[sin∅cos¶ + cos∅sin¶. -sin∅sin¶ +cos∅cos¶

hence AB=BA