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प्रश्न
If `sec θ = 13/5`, show that `(2 sin θ - 3 cos θ)/(4 sin θ - 9 cos θ) = 3`.
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उत्तर १
Given: `sec θ = 13/5`
We know that,
sec θ = `"Hypotenuse"/"Adjacent Side"`
sec θ = `13/5 = "AC"/"BC"`
Let AC = 13k and BC = 5k
In ΔABC, ∠B = 90°
By Pythagoras theorem,
AC2 = AB2 + BC2
(13k)2 = AB2 + (5k)2
AB2 = 169k2 – 25k2
AB2 = 144k2
AB = 12k
sin θ = `"AB"/"AC" = "12k"/"13k" = 12/13`
cos θ = `"BC"/"AC" = "5k"/"13k" = 5/13`
LHS = `(2 sin θ - 3 cos θ)/(4 sin θ - 9 cos θ)`
LHS = `[2 × (12/13) - 3 × (5/13)]/[4 × (12/13) - 9 × (5/13)]`
LHS = `[24/13 - 15/13]/[48/13 + 45/13]`
LHS = `[9/13]/[3/13]`
LHS = `9/(cancel13) × cancel13/3`
LHS = `9/3`
LHS = 3
RHS = 3
LHS = RHS
`(2 sin θ - 3 cos θ)/(4 sin θ - 9 cos θ) = 3`
Hence proved.
उत्तर २
Given: sec θ = `13/5`
cos θ = `1/secθ = 5/13`
sin2θ = 1 – cos2θ
sin2θ = `1 - (5/13)^2`
sin2θ = `1 - 25/169`
sin2θ = `(169 − 25)/169`
sin2θ = `144/169`
sin θ = `12/13`
Now, put the values in the equation,
LHS = `(2 sin θ - 3 cos θ)/(4 sin θ - 9 cos θ)`
LHS = `(2 × (12/13) - 3 × (5/13))/(4 × (12/13) - 9 × (5/13))`
LHS = `(24/13 - 15/13)/(48/13 - 45/13)`
LHS = `((24- 15)/cancel13)/((48 - 45)/cancel13)`
LHS = `9/3`
LHS = 3
RHS = 3
LHS = RHS
`(2 sin θ - 3 cos θ)/(4 sin θ - 9 cos θ) = 3`
Hence proved.
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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)`
= `((1 + sin θ) square)/(cos θ square)` ......[Multiplying `square` with the numerator and denominator]
= `(1^2 - square)/(cos θ square)`
= `square/(cos θ square)`
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∴ L.H.S. = R.H.S.
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