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Questions
Prove the following trigonometric identities.
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
Prove that:
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
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Solution
We have to prove `(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
We know that, sin2 θ + cos2 θ = 1
Multiplying both numerator and denominator by (1 − sin θ), we have
`(1 - sin θ)/(1 + sin θ) = ((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`
`= (1 - sin θ)^2/(1 - sin^2 θ)`
`= ((1 - sin θ)/cos θ)^2`
`= (1/cos θ - sin θ/cos θ)^2`
`= (sec θ - tan θ)^2`
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
