In the given figure; ∠C = 90° and D is mid-point of AC. Find :

(i) $\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}}$

(ii) $\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}}$

Since D is the mid-point of A. So, AC = 2DC

(i) $\text{tan ∠CAB} = \dfrac{Perpendicular}{Base} = \dfrac{BC}{AC}$

$\text{tan ∠CDB} = \dfrac{Perpendicular}{Base} = \dfrac{BC}{DC}$

Now,

$\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}}\\[1em] = \dfrac{\dfrac{BC}{AC}}{\dfrac{BC}{DC}}\\[1em] = \dfrac{BC \times DC}{AC \times BC}\\[1em] = \dfrac{\cancel{BC} \times DC}{AC \times \cancel{BC}}\\[1em] = \dfrac{DC}{AC}\\[1em] = \dfrac{DC}{2 \times DC}\\[1em] = \dfrac{\cancel{DC}}{2 \times \cancel{DC}}\\[1em] = \dfrac{1}{2}$

Hence, $\dfrac{\text{tan ∠CAB}}{\text{tan ∠CDB}} = \dfrac{1}{2}$

(ii) $\text{tan ∠ABC} = \dfrac{Perpendicular}{Base} = \dfrac{AC}{BC}$

$\text{tan ∠DBC} = \dfrac{Perpendicular}{Base} = \dfrac{DC}{BC}$

Now,

$\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}}\\[1em] = \dfrac{\dfrac{AC}{BC}}{\dfrac{DC}{BC}}\\[1em] = \dfrac{AC \times BC}{BC \times DC}\\[1em] = \dfrac{AC \times \cancel{BC}}{\cancel{BC} \times DC}\\[1em] = \dfrac{AC}{DC}\\[1em] = \dfrac{2 \times DC}{DC}\\[1em] = \dfrac{2 \times \cancel{DC}}{\cancel{DC}}\\[1em] = 2$

Hence, $\dfrac{\text {tan ∠ABC}}{\text {tan ∠DBC}} = 2$