The trick here is to recall that 1 = cos^a+sin^2A
cos2A/(1+sin2A)
(cos^2A-sin^2A)/(1+2sinAcosA)
(cos^2A-sin^2A)/(cos^2A+2sinAcosA+sin^2A)
(cosA+sinA)(cosA-sinA)/(cosA+sinA)^2
(cosA-sinA)/(cosA+sinA)
(cotA-1)/(cotA+1)
Prove that (cos2A)÷ ( 1+sin2A) = (cotA -1) ÷ (cotA +1)
2 answers
The trick here is to recall that 1 = cos^a+sin^2A
cos2A/(1+sin2A)
(cos^2A-sin^2A)/(1+2sinAcosA)
(cos^2A-sin^2A)/(cos^2A+2sinAcosA+sin^2A)
(cosA+sinA)(cosA-sinA)/(cosA+sinA)^2
(cosA-sinA)/(cosA+sinA)
(cotA-1)/(cotA+1)
cos2A/(1+sin2A)
(cos^2A-sin^2A)/(1+2sinAcosA)
(cos^2A-sin^2A)/(cos^2A+2sinAcosA+sin^2A)
(cosA+sinA)(cosA-sinA)/(cosA+sinA)^2
(cosA-sinA)/(cosA+sinA)
(cotA-1)/(cotA+1)