tan²α=2tan²θ+1求证cos2α+sin²θ=0

问题描述:

tan²α=2tan²θ+1求证cos2α+sin²θ=0

  ∵tan²α=2tan²θ+1
  ∴tan²α=-1
  ∴cos2α+sin²θ
  =cos²a-sin²+sin²
  =cos²a
  =1/[sec²a]
  =1/[1+tan²a]
  =1/0
  =∞