三角比证明(secα)^2*(tanβ)^2-(tanα)^2*(secβ)^2=(tanβ)^2-(tanα)^2

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三角比证明(secα)^2*(tanβ)^2-(tanα)^2*(secβ)^2=(tanβ)^2-(tanα)^2
三角比证明
(secα)^2*(tanβ)^2-(tanα)^2*(secβ)^2=(tanβ)^2-(tanα)^2

三角比证明(secα)^2*(tanβ)^2-(tanα)^2*(secβ)^2=(tanβ)^2-(tanα)^2
左边=[(tana)^2+1](tanb)^2-(tana)^2[(tanb)^2+1]
=(tana)^2(tanb)^2+(tanb)^2-(tana)^2(tanb)^2-(tana)^2
=(tanb)^2-(tana)^2
=右边

(secα)^2*(tanβ)^2-(tanα)^2*(secβ)^2
=(1+tan^2 α)*(tan β)^2-(tan α )^2*(1+tan^2 β)
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=(tanβ)^2-(tanα)^2