\[A+iB\] form of \[\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(1+i\tan \,\,v)}\]is equal to:

\[\sin \,u\,\cos \,v[\cos \,(x+y+u+v)+\]\[i\sin (x+y+u+v)]\]

\[\sin u\,\,\cos v\,\,[\cos (x+y-u-v)+\]\[i\sin (x+y-u-v)]\]

\[\sin \,\,u\,\,\,\cos \,\,v\,\,[\cos \,(x+y+u+v)-\]\[i\sin (x+y-u+v)]\]

\[A+iB\] form of \[\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(

1.	\[A+iB\] form of \[\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(1+i\tan \,\,v)}\]is equal to:
A.	\[\sin u\,\,\cos v\,\,[\cos (x+y-u-v)+\]\[i\sin (x+y-u-v)]\]
B.	\[\sin \,u\,\cos \,v[\cos \,(x+y+u+v)+\]\[i\sin (x+y+u+v)]\]
C.	\[\sin \,\,u\,\,\,\cos \,\,v\,\,[\cos \,(x+y+u+v)-\]\[i\sin (x+y-u+v)]\]
D.	None of these
Answer» B. \[\sin \,u\,\cos \,v[\cos \,(x+y+u+v)+\]\[i\sin (x+y+u+v)]\]

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\[A+iB\] form of \[\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(1+i\tan \,\,v)}\]is equal to:

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