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Question
Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.
Proof: L.H.S. = sec θ + tan θ
= `1/square + square/square`
= `square/square` ......`(∵ sec θ = 1/square, tan θ = square/square)`
= `((1 + sin θ) square)/(cos θ square)` ......[Multiplying `square` with the numerator and denominator]
= `(1^2 - square)/(cos θ square)`
= `square/(cos θ square)`
= `cos θ/(1 - sin θ)` = R.H.S.
∴ L.H.S. = R.H.S.
∴ sec θ + tan θ = `cos θ/(1 - sin θ)`
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Solution
Proof: L.H.S. = sec θ + tan θ
= `1/bb(cos θ) + bb(sin θ)/bb(cos θ)` ........`[∵ sec θ = 1/bb(cos θ), tan θ = bb(sin θ)/bb(cos θ)]`
= `bb(1 + sintheta)/bbcostheta` = `((1 + sin θ) bb(1 - sin θ))/(cos θ bb(1 - sin θ)` ......[Multiplying `bb(1 - sin θ)` with the numerator and denominator]
= `(1^2 - bb(sin^2 θ))/(cos θ bb(1 - sin θ)`
= `bb (cos^2 θ)/(cos θ bb(1 - sin θ)`
= `cos θ/(1 - sin θ)` = R.H.S.
∴ L.H.S. = R.H.S.
∴ sec θ + tan θ = `cos θ/(1 - sin θ)`
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