Verify the trig identity: (1-sinΘ)/(1+sinΘ) = (secΘ-tanΘ)²
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Verify the trig identity: (1-sinΘ)/(1+sinΘ) = (secΘ-tanΘ)²

[From: ] [author: ] [Date: 11-12-23] [Hit: ]
..or quotient identities sec θ = 1/cos θ, tan θ = sin θ / cos θ.-1) Multiplying both nr. and dr.......
Help! Explain and show the steps!

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Start with the left side, and multiply by 1 = (1 - sin θ)/(1 - sin θ) to get

(1 - sin θ)/(1 + sin θ) = (1 - sin θ)²/(1 - sin² θ)
= (1 - sin θ)² / cos² θ
= [ (1 - sin θ) / cos θ ]²
= (sec θ - tan θ)²

I used the product of sum and difference shortcut to get a difference of squares in the first line's denominator. The last line just applies the definition of the tan and sec functions...or "quotient identities" sec θ = 1/cos θ, tan θ = sin θ / cos θ.

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1) Multiplying both nr. and dr. by {1 - sin(θ)},

Left side = {1 - sin(θ)}²/(1 - sin²θ) [since (a+b)(a-b) = a² - b²]

= {1 - sin(θ)}²/(cos²θ)

= {1/cos(θ) - sin(θ)/cos(θ)}² = {sec(θ) - tan(θ)}² [Proved]

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Two things are infinite : the universe and human stupidity; Im not sure about the universe.
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