Question

If sec θ = `13/5, "show that"  (2sinθ - 3 cosθ)/(4sinθ - 9cosθ) = 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 Pathagoras 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 = `(2sinθ - 3 cosθ)/(4sinθ - 9cosθ)`

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

`(2sinθ - 3 cosθ)/(4sinθ - 9cosθ) = 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 = `(2sinθ - 3 cosθ)/(4sinθ - 9cosθ)`

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

`(2sinθ - 3 cosθ)/(4sinθ - 9cosθ) = 3`

Hence proved.