1/1-sinα+1/1+sinα已知tanα=2

详细点

1/(1-sinα)+1/(1+sinα)
=(1+sinα)/[(1-sinα)(1+sinα)]+(1-sinα)/[(1-sinα)(1+sinα)]
=(1+sinα)/(1-sin^2α)+(1-sinα)/(1-sin^2α)
=(1+sinα+1-sinα)/(1-sin^2α)
=2/cos^2α
=2(sin^2α+cos^2α)/cos^2α
=2(sin^2α+cos^2α)/cos^2α 分子分母同时同时除以cos^2α
=2(sin^2α/cos^2α+cos^2α/cos^2α)/(cos^2α/cos^2α)
=2(tan^2α+1)/1
=2(tan^2α+1)
=2*(2^2+1)
=2*5
=10

解：∵tana=2 ∴1+tan²a=5 ∵1+tan²a=sec²a ∴sec²a=5 ∴1/(1-sina)+1/(1+sina)=(1+sina+1-sina)/(1-sin²a) =2/cos²a =2sec²a =10

sinθ(1 tanθ) cosθ(1 1/tanθ) = (sinθ/cosθ)(cosθ sinθ) (cosθ/sinθ)(sinθ cosθ) = (sinθ cosθ)(sinθ/cosθ cosθ/sinθ) = (sinθ cosθ)(sin²θ cos²θ)/(sinθcosθ) = 1/cosθ 1/sinθ = 左边 证毕！