If SecA +TanA=m then show that (m²-1)/(m²+1)=SinA

Pls help me

1) Given m = sec(A) + tan(A) = [1 + sin(A)]/cos(A)

2) So, m² = {1 + sin(A)}²/cos²A = {1 + sin(A)}²/(1 - sin²A) [Since cos²A + sin²A =1]
==> m² = {1 + sin(A)}²/{1 + sin(A)}*{1 - sin(A)} = {1 + sin(A)}/{1 - sin(A)}

3) So, {(m² - 1)/(m² + 1)} = [{1 + sin(A) - 1 + sin(A)}cos²A]/[{1 + sin(A) + 1 - sin(A)]
[Applying compenendo-dividendo property, that is if a/b = c/d,
then (a-b)/(a+b) = (c-d)/(c+d)]

4) (m² - 1)/(m² + 1) = 2sin(A)/2 = sin(A) = Right side [Proved]

sec(A) + tan(A) = 1/cos(A) + sin(A)/cos(A) = (1+sin(A))/cos(A) = m

[((1+sin(A))/cos(A))^2 - 1)]/[((1+sin(A))/cos(A))^2 + 1)]

[(1 + 2sin(A) + sin^2(A) - cos^(A))/cos^2(A)]/[(1 + 2sin(A) + sin^2(A) + cos^2(A))/cos^2(A)]

[(2sin(A) + 2sin^2(A))/cos^2(A)]/[(2 + 2sin(A))/cos^2(A)]

(2sin(A) + 2sin^2(A))/(2 + 2sin(A)) -----> 2sin(A)[1 + sin(A)]/(2[1 + sin(A)]) -----> 2sin(A)/2 -----> sin(A)

m = secA + tanA = 1/cosA + sinA/cosA = (1 + sinA)/cosA
m² = (1 + sinA)² / cos²A = (1 + 2sinA + sin²A) / cos²A
m² - 1 = [(1 + 2sinA + sin²A) / cos²A] - 1 = (1 + 2sinA + sin²A - cos²A) / cos²A
m² + 1 = [(1 + 2sinA + sin²A) / cos²A] + 1 = (1 + 2sinA + sin²A + cos²A) / cos²A

(m² - 1) / (m² + 1)
= (1 + 2sinA + sin²A - cos²A) / cos²A / (1 + 2sinA + sin²A + cos²A) / cos²A
= (1 + 2sinA + sin²A - cos²A) / (1 + 2sinA + sin²A + cos²A)
= (1 + 2sinA + 2sin²A - 1) / (1 + 2sinA + 1)
= 2sinA(1 + sinA) / 2(1 + sinA)
= sinA

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