Assertion :(cosA+cosBsinA−sinB)n+(sinA+sinBcosA−cosB)n
=2cotn(A−B2) if n is even
=0 if n is odd. Reason: cosA+cosBsinA−sinB=cotA−B2.
cosA+cosBsinA−sinB=2cos(A+B2)cos(A−B2)2cos(A+B2)sin(A−B2)=cot(A−B2)
sinA+sinBcosA−cosB=2sin(A+B2)cos(A−B2)−2sin(A+B2)sin(A−B2)=−cot(A−B2)
(cosA+cosBsinA−sinB)n+(sinA+sinBcosA−cosB)n=(cot(A−B2))n+(−cot(A−B2))n
If n is even
(cosA+cosBsinA−sinB)n+(sinA+sinBcosA−cosB)n=2cot(A−B2)
Hence, option 'A' is correct.