Question

已知tanB+cotB=5,求sinB+cosB的值。

写出解题过程

Answer 1

tanB+cotB=5
即sinB/cosB+cosB/sinB=5
通分有 [(sinB)^2+(cosB)^2]/sinBcosB=5
即1/sinBcosB=5
所以sinBcosB=1/5
而（sinB+cosB)^2=(sinB)^2+(cosB)^2+2sinBcosB
=1+2sinBcosB
=7/5
所以sinB+cosB=根号（7/5)

Answer 2

sinB/cosB+cosB/sinB=5,
1/sinB*cosB=5,
sinB*cosB=1/5.
(sinB+cosB)^2=1+2sinB*cosB=1+2/5=7/5,
sinB+cosB=根号35/5,(负值舍去了)

Answer 3

sinB/cosB+cosB/sinB
=[(sinB)^2+(cosB)^2]/(sinbcosB)
=1/(sinBcosB)=5
sinBcosB=1/5

(sinB+cosB)^2=1+2sinBcosB=1+2/5=7/5

sinB+cosB=±√(7/5)