tan²α=2tan²θ+1求证cos2α+sin²θ=0
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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
=∞
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
=∞