If cos a=1/k and 0<a<90, then tan(a-270)=?

2016-09-03 7:17 am

回答 (1)

2016-11-12 4:11 am
cos a=1/k
(sin a)^2 +(cos a)^2=1
sin a=[1-(1/k)^0.5]^0.5
tan(a-270)
=sin(a-270)/cos(a-270)
=[(sin a)(cos 270)-(cos a)(sin 270)]/[(cos a)(cos 270)+(sin a)(sin 270)]
=(cos a)/(-sin a)
=-(1/k)/[1-(1/k)^2]^0.5
=-(1/k)/[(k^2 -1)/k^2]^0.5
=-1/(k^2 -1)^0.5


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