1) limx→π/4tanx−1cos2x is equal to A) 1 B) 0 C) -2 D) -1 Answer: Option DExplanation:limx→π/4tanx−1cos2x=limh→0tan(π4+h)−1cos2(π4+h) [∵ x=π4+h] =limh→0(1+tanh1−tanh)−1cos(π2+2h) =limh→01+tanh−1+tanh−sin2h(1−tanh) = limh→0−2tanh2sinhcosh(1−tanh) = limh→0−1cos2h(1−tanh)=−1