how is (secx + tanx)^2 equal to (1 + sinx)/(1 - sinx) ?

L.H.S.
= (secx + tanx)²
= [(1/cosx) + (sinx/cosx)]² ...... as secx = 1/cosx and tanx = sinx/cosx
= [(1+ sinx)/cosx]²
= (1+ sinx)²/cos²x
= (1+ sinx)²/(1 - sin²x) ...... as cos²x + sin²x = 1
= (1+ sinx)²/[(1 + sinx)(1 - sinx)] ...... as 1 - sin²x = (1 + sinx)(1 - sinx)
= (1 + sinx)/(1 - sinx) ...... eliminate (1 - sinx) in denominator and numerator, provided (1 - sinx) ≠ 0
= R.H.S.

Hence, (secx + tanx)² = (1 + sinx)/(1 - sinx)

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(secx + tanx)^2 = (1 + sinx)/(1 - sinx),
L.H.S. = (secx + tanx)^2 = (1/cosx + sinx/cosx)² = ((1+sinx)/cosx)² =
= (1+sinx)²/cos²x = (1+sinx)²/(1−sin²x) = (1+sinx)²/((1−sinx)(1+sinx)) =
= (1+sinx)/(1−sinx) = R.H.S. ,_________________/ Q.E.D.

(sec x + tan x)^2
= (1 + sin x)^2/cos^2 x
= (1 + sin^2 x + 2 sin x ) /(1 - sin^2 x)
= (2 sin^2 x + cos^2 x + 2 sin x)/(1 - sin^2 x)
= 2 sin x(sin x + 1) + cos^2 x / (1 - sin^2 x)
= 2 sin x(sin x + 1) + 1 - sin^2 x / (1 - sin^2 x)
= (sin^2 x + 2 sin x + 1) / (1 - sin^2 x)
= (1 + sin x)(1 + sin x) / (1 + sin x)(1 - sin x)
= (1 + sin x) / (1 - sin x)

= [sec(x) + tan(x)]² → you know that: sec(x) = 1/cos(x)

= [{1/cos(x)} + tan(x)]² → you know that: tan(x) = sin(x)/cos(x)

= [{1/cos(x)} + {sin(x)/cos(x)}]²

= [1/cos(x)]² + 2.[1/cos(x)].[sin(x)/cos(x)] + [sin(x)/cos(x)]²

= [1/cos²(x)] + 2.[sin(x)/cos²(x)] + [sin²(x)/cos²(x)]

= [1 + 2.sin(x) + sin²(x)]/cos²(x)

= [1 + sin(x)]²/cos²(x) → recall: cos²(x) + sin²(x) = 1 → cos²(x) = 1 - sin²(x)

= [1 + sin(x)]²/[1 - sin²(x)] → recall: a² - b² = (a + b).(a - b)

= [1 + sin(x)]²/ { [1 + sin(x)].[1 - sin(x)] } → you can simplify by [1 - sin(x)]

= [1 + sin(x)]/[1 - sin(x)]

　
(secx + tanx)²
= (1/cosx + sinx/cosx)²
= ((1+sinx)/cosx)²
= (1+sinx)²/cos²x
= (1+sinx)²/(1−sin²x)
= (1+sinx)²/((1−sinx)(1+sinx))
= (1+sinx)/(1−sinx)