Given : 17 cos θ = 15; find the value of tan θ + 2 sec θ.

Given:

17 cos θ = 15

cos θ = $\dfrac{15}{17}$

$\Rightarrow \text{cos θ} = \dfrac{Base}{Hypotenuse} = \dfrac{15}{17}\\[1em]$

∴ If length of AC = 17x unit, length of AB = 15x unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ (17x)² = BC² + (15x)²

⇒ 289x² = BC² + 225x²

⇒ BC² = 289x² - 225x²

⇒ BC = $\sqrt{64 \text{x}^2}$

⇒ BC = 8x

tan θ = $\dfrac{Perpendicular}{Base}$

$= \dfrac{CB}{AB} = \dfrac{8x}{15x} = \dfrac{8}{15}$

sec θ = $\dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{AB} = \dfrac{17x}{15x} = \dfrac{17}{15}$

Now, tan θ + 2 sec θ

$= \dfrac{8}{15} + 2 \times \dfrac{17}{15}\\[1em] = \dfrac{8}{15} + \dfrac{34}{15}\\[1em] = \dfrac{8 + 34}{15}\\[1em] = \dfrac{42}{15}\\[1em] = \dfrac{14}{5}\\[1em] = 2\dfrac{4}{5}$

Hence, tan θ + 2 sec θ = $2\dfrac{4}{5}$.