The vertices of a triangle are (0, 0),(sec θ, tanθ) and (4, -5) (0<θ<∏/2). The minimum possible area is --------

Dear ANAM

open determinant

Δ =1/2| [-5secΘ -4tanΘ]|

Δ =1/2| [5secΘ +4tanΘ]|

  clearly at Θ =0 both tanΘ  and sec Θ is minimum in there given domain.

so put Θ=0

Δ =5/2

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