Given : $\text{cos A} = \dfrac{5}{13}$

evaluate :

(i) $\dfrac{\text{sin A} - \text{cot A}}{2 \text{ tan A}}$

(ii) $\text{cot A} + \dfrac{1}{\text{cos A}}$

Given:

cos $A = \dfrac{5}{13}$

i.e. $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{5}{13}$

∴ If length of BA = 5x unit, length of AC = 13x unit.

In Δ ABC,

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

⇒ (13x)² = (5x)² + BC²

⇒ 169x² = 25x² + BC²

⇒ BC² = 169x² - 25x²

⇒ BC² = 144x²

⇒ BC = $\sqrt{144\text{x}^2}$

⇒ BC = 12x

(i) sin $A = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{AC} = \dfrac{12x}{13x} = \dfrac{12}{13}$

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

$= \dfrac{AB}{BC} = \dfrac{5x}{12x} = \dfrac{5}{12}$

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

$= \dfrac{BC}{AB} = \dfrac{12x}{5x} = \dfrac{12}{5}$

Now,

$= \dfrac{\text{sin A} - \text{cot A}}{2 \text{ tan A}}\\[1em] = \dfrac{\dfrac{12}{13} - \dfrac{5}{12}}{2 \times \dfrac{12}{5}}\\[1em] = \dfrac{\dfrac{12 \times 12}{13 \times 12} - \dfrac{5 \times 13}{12 \times 13}}{\dfrac{24}{5}}\\[1em] = \dfrac{\dfrac{144}{156} - \dfrac{65}{156}}{\dfrac{24}{5}}\\[1em] = \dfrac{\dfrac{144 - 65}{156}}{\dfrac{24}{5}}\\[1em] = \dfrac{\dfrac{79}{156}}{\dfrac{24}{5}}\\[1em] = \dfrac{79 \times 5}{156 \times 24}\\[1em] = \dfrac{395}{3,744}$

Hence, $\dfrac{\text{sin A} - \text{cot A}}{2 \text{ tan A}} = \dfrac{395}{3,744}$.

(ii) cos $A = \dfrac{5}{13}$

cot $A = \dfrac{5}{12}$

To find,

$\text{cot A} + \dfrac{1}{\text{cos A}}$

$\text{cot A} + \dfrac{1}{\text{cos A}} = \dfrac{5}{12} + \dfrac{1}{\dfrac{5}{13}}\\[1em] = \dfrac{5}{12} + \dfrac{13}{5}\\[1em] = \dfrac{5 \times 5}{12 \times 5} + \dfrac{13 \times 12}{5 \times 12}\\[1em] = \dfrac{25}{60} + \dfrac{156}{60}\\[1em] = \dfrac{25 + 156}{60}\\[1em] = \dfrac{181}{60}$

Hence, $\text{cot A} + \dfrac{1}{\text{cos A}} = \dfrac{181}{60}$.