From the following figure, find the values of :

(i) cos A

(ii) cosec A

(iii) tan²A – sec²A

(iv) sin C

(v) sec C

(v) cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}}$

In Δ ADB,

⇒ AB² = AD² + BD² (∵ AB is hypotenuse)

⇒ AB² = (3)² + (4)²

⇒ AB² = 9 + 16

⇒ AB² = 25

⇒ AB = $\sqrt{25}$

⇒ AB = 5

In Δ CDB,

⇒ BC² = BD² + DC² (∵ BC is hypotenuse)

⇒ (12)² = (4)² + DC²

⇒ 144 = 16 + DC²

⇒ DC² = 144 - 16

⇒ DC² = 128

⇒ DC = $\sqrt{128}$

⇒ DC = $8\sqrt{2}$

(i) cos $A = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AD}{AB}\\[1em] = \dfrac{3}{5}$

Hence, cos $A = \dfrac{3}{5}$.

(ii) cosec $A = \dfrac{Hypotenuse}{Perpendicular}$

$= \dfrac{AD}{BD}\\[1em] = \dfrac{5}{4}$

Hence, cosec $A = \dfrac{5}{4}$.

(iii) tan²A – sec²A

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

$= \dfrac{BD}{AD}\\[1em] = \dfrac{4}{3}$

sec $A = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AB}{AD}\\[1em] = \dfrac{5}{3}$

tan²A – sec²A

$= \Big(\dfrac{4}{3}\Big)^2 - \Big(\dfrac{5}{3}\Big)^2\\[1em] = \Big(\dfrac{16}{9}\Big) - \Big(\dfrac{25}{9}\Big)\\[1em] = \Big(\dfrac{16 - 25}{9}\Big)\\[1em] = \Big(\dfrac{-9}{9}\Big)\\[1em] = -1$

Hence, tan²A – sec²A = -1.

(iv) sin $C = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BD}{BC}\\[1em] = \dfrac{4}{12}\\[1em] = \dfrac{1}{3}\\[1em]$

Hence, sin $C = \dfrac{1}{3}$.

(v) sec $C = \dfrac{Hypotenuse}{Base}$

$= \dfrac{BC}{DC}\\[1em] = \dfrac{12}{8 \sqrt{2}}\\[1em] = \dfrac{3}{2 \sqrt{2}}\\[1em] = \dfrac{3 \times \sqrt{2}}{2 \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{3 \sqrt{2}}{2 \times 2}\\[1em] = \dfrac{3 \sqrt{2}}{4}$

Hence, sec $C = \dfrac{3 \sqrt{2}}{4}$.

(vi) cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}}$

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

$= \dfrac{Base}{Perpendicular}\\[1em] = \dfrac{8 \sqrt{2}}{4}\\[1em] = 2 \sqrt{2}$

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

$= \dfrac{4}{12}\\[1em] = \dfrac{1}{3}$

Now, cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}}$

$= (2 \sqrt{2})^2 - \dfrac{1}{\Big(\dfrac{1}{3}\Big)^2}\\[1em] = (2 \sqrt{2})^2 - \dfrac{9}{1}\\[1em] = 4 \times 2 - 9\\[1em] = 8 - 9\\[1em] = -1$

Hence, cot²C - $\dfrac{1}{\text{sin}^2 \text{ C}} = -1$.