Question
prove that 2sinA + 1 / cosA+sin2A=secA
Answers
Steve
(2sinA+1)/(cosA+sin2A)
= (2sinA+1)/(cosA+2sinAcosA)
= (2sinA+1)/[cosA(1+2sinA)]
= 1/cosA
= secA
= (2sinA+1)/(cosA+2sinAcosA)
= (2sinA+1)/[cosA(1+2sinA)]
= 1/cosA
= secA