神马作文网若tan(a/2+π/4)=3+2·sqrt2,则(1-sina)/cosa的值是?
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若tan(a/2+π/4)=3+2·sqrt2,则(1-sina)/cosa的值是?
A.3+2·sqrt2 B.3-2·sqrt2 C.17+12·sqrt2 D.17-12·sqrt2
A.3+2·sqrt2 B.3-2·sqrt2 C.17+12·sqrt2 D.17-12·sqrt2
tan(a/2+π/4)
=[tan(a/2)+tan(π/4)]/1-tan(a/2)tan(π/4)
=[tan(a/2)+1]/1-tan(a/2)--------------(1)
=3+2√2
(1-sina)/cosa
=[(sina/2)^2+(cosa/2)^2-2sina/2cosa/2]/(cosa/2)^2-(sina/2)^2
=[cosa/2-sina/2]/[cosa/2+sina/2]
=1/(1)
=3-2√2
B.3-2·sqrt2
=[tan(a/2)+tan(π/4)]/1-tan(a/2)tan(π/4)
=[tan(a/2)+1]/1-tan(a/2)--------------(1)
=3+2√2
(1-sina)/cosa
=[(sina/2)^2+(cosa/2)^2-2sina/2cosa/2]/(cosa/2)^2-(sina/2)^2
=[cosa/2-sina/2]/[cosa/2+sina/2]
=1/(1)
=3-2√2
B.3-2·sqrt2
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