Given : $\text{sec A} =\dfrac{29}{21}$, evaluate : $\text{sin A} - \dfrac{1}{\text{tan A}}$

Given:

sec $A = \dfrac{29}{21}$

i.e. $\dfrac{\text{Hypotenuse}}{\text{Base}} = \dfrac{29}{21}$

∴ If length of AB = 21x unit, length of AC = 29x unit.

In Δ ABC,

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

⇒ (29x)² = (21x)² + BC²

⇒ 841x² = 441x² + BC²

⇒ BC² = 841x² - 441x²

⇒ BC² = 400x²

⇒ BC = $\sqrt{400\text{x}^2}$

⇒ BC = 20x

sin $A = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{AC} = \dfrac{20x}{29x} = \dfrac{20}{29}$

tan $A = \dfrac{Perpendicular}{Base}$

$= \dfrac{BC}{AB} = \dfrac{20x}{21x} = \dfrac{20}{21}$

Now,

$\text{sin A} - \dfrac{1}{\text{tan A}}\\[1em] = \dfrac{20}{29} - \dfrac{1}{\dfrac{20}{21}}\\[1em] = \dfrac{20}{29} - \dfrac{21}{20}\\[1em] = \dfrac{20 \times 20}{29 \times 20} - \dfrac{21 \times 29}{20 \times 29}\\[1em] = \dfrac{400}{580} - \dfrac{609}{580}\\[1em] = \dfrac{400 - 609}{580}\\[1em] = \dfrac{- 209}{580}\\[1em]$

Hence, $\text{sin A} - \dfrac{1}{\text{tan A}} = \dfrac{- 209}{580}$.