prove that

Asked by  | 6th Jun, 2008, 07:30: PM

Expert Answer:

tanA/(secA-1) + tanA/(secA+1)

= (sinA/cosA)/(1-cosA / cosA) + (sinA/cosA)/(1+cosA/cosA)

= sinA/(1-cosA) + sinA/(1+cosA)

= [sinA - sinAcosA + sinA + sinAcosA] / (1+cosA)(1-cosA)

= 2sinA/ 1-cos2A

= 2sinA/ sin2A

= 2/sinA

= 2CosecA

Answered by  | 7th Jun, 2008, 08:18: PM

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