Question

If `sec θ = 13/5`, show that `(2 sin θ - 3 cos θ)/(4 sin θ - 9 cos θ) = 3`.

Answer 1

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.

Answer 2

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.