TanB+ cotB = secB cscB

Question

It's proving identities ( often found in trigonometry of advanced functions)

Answer 1

TanB+ cotB = secB cscB

sinB/cosB+cosB/sinB=(sin²B+cos²B)/(sinBcosB)=1/(sinBcosB)=cscBsecB

Answer 2

tanB + cotB = (sinB)/(cosB) + (cosB)/(sinB)

 

                      = (sin²B + cos²B)/[(cosB)(sinB)]

 

                       = 1/[(cosB)(sinB)]

 

                        = (1/cosB)(1/sinB)

 

                        = (secB)(cscB)