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Question
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
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Solution
`(secθ - tanθ)^2`
= `(1/cosθ - sinθ/cosθ)^2`
= `((1 - sinθ)/cosθ)^2 = (1 - sinθ)^2/cos^2θ`
= `(1 - sinθ)^2/(1 - sin^2θ) = (1 - sinθ)^2/((1 -sinθ)(1 + sinθ))` (∵ `1 - sin^2θ = cos^2θ`
= `(1 - sinθ)/(1 + sinθ)`
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Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
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`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
