how would you solve this. it cant be solved by L'hospital's rule
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lim{x→π/2} [1/cos(x)] / [sin(x)/cos(x)]
= lim{x→π/2} csc(x) = 1
= lim{x→π/2} csc(x) = 1
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lim_{x→π/2} sec(x)/tan(x)
= lim_{x→π/2} (1/cos(x)) / (sin(x)/cos(x))
= lim_{x→π/2} cos(x) / (cos(x)sin(x))
= lim_{x→π/2} 1/sin(x)
= 1
= lim_{x→π/2} (1/cos(x)) / (sin(x)/cos(x))
= lim_{x→π/2} cos(x) / (cos(x)sin(x))
= lim_{x→π/2} 1/sin(x)
= 1