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c o t ( 2 π − ζ ) = t a n ( ζ ) , hence the question may be written as ( 1 + c o s e c ( θ ) s e c 2 ( ζ ) − t a n 2 ( ζ ) ) ( 1 − s e c ( θ ) t a n ( θ ) ) or in a more simplified form as ( 1 + s i n ( θ ) ) ( 1 − s i n ( θ ) ) which may be put further to 1 − s i n 2 ( θ ) , an equivalent to ∣ c o s ( θ ) ∣ . Given that 0 < θ < 2 π , the value may be released from the absolute value as c o s ( θ ) = 7 2 , hence s e c ( θ ) = 2 7