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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) - Geometry Mathematics 2

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प्रश्न

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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उत्तर

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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