If sec θ = $\dfrac{13}{5}$, show that $\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}}$ = 3.

sec θ = $\dfrac{\text{hypotenuse}}{\text{base}} =\dfrac{13}{5}$

Let hypotenuse = 13x and base = 5x

We will find perpendicular by using pythagoras theorem

Hypotenuse² = Base² + Perpendicular²

Perpendicular² = Hypotenuse² - Base²

Perpendicular² = (13x)² - (5x)²

Perpendicular² = 169x² - 25x²

Perpendicular² = 144x²

Perpendicular = $\sqrt{144x^2}$

Perpendicular = 12x

Now

sin θ = $\dfrac{\text{perpendicular}}{\text{hypotenuse}}= \dfrac{12x}{13x} = \dfrac{12}{13}$

cos θ = $\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{5x}{13x} = \dfrac{5}{13}$

Substituting values we get :

$\Rightarrow \dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = \dfrac{2 \times \dfrac{12}{13} - 3 \times \dfrac{5}{13}}{4 \times \dfrac{12}{13} - 9 \times \dfrac{5}{13}} \\[1em] = \dfrac{\dfrac{24}{13} - \dfrac{15}{13}}{\dfrac{48}{13} - \dfrac{45}{13}} \\[1em] = \dfrac{\dfrac{9}{13}}{\dfrac{3}{13}} \\[1em] = \dfrac{9}{3} \\[1em] = 3.$

Hence, proved that $\dfrac{\text{2 sin θ - 3 cos θ}}{\text{4 sin θ - 9 cos θ}} = 3$.