In the following figure :

AD ⊥ BC, AC = 26, CD = 10, BC = 42, ∠DAC = x and ∠B = y.

Find the value of :

(i) cot x

(ii) $\dfrac{1}{\text{ sin}^2 \text{ y}} -\dfrac{1}{\text{tan}^2 \text{y}}$

(iii) $\dfrac{6}{\text{ cos x}} -\dfrac{5}{\text{ cos y}} + 8 {\text{ tan y}}$.

Given,

BC = 42 and CD = 10

From figure,

BD = BC - CD

⇒ BD = 42 - 10 = 32

In Δ ACD,

⇒ AC² = AD² + CD² (∵ AC is hypotenuse)

⇒ (26)² = AD² + (10)²

⇒ 676 = AD² + 100

⇒ AD² = 676 - 100

⇒ AD² = 576

⇒ AD = $\sqrt{576}$

⇒ AD = 24

In Δ ADB,

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

⇒ AB² = (24)² + (32)²

⇒ AB² = 576 + 1,024

⇒ AB² = 1,600

⇒ AB = $\sqrt{1,600}$

⇒ AB = 40

(i) cot $x = \dfrac{Base}{Perpendicular}$

$= \dfrac{AD}{CD} = \dfrac{24}{10} = 2.4$

(ii) sin $y = \dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{AD}{AB} = \dfrac{24}{40} = \dfrac{3}{5}$

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

$= \dfrac{AD}{DB} = \dfrac{24}{32} = \dfrac{3}{4}$

Now,

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

(iii) cos $x = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AD}{AC} = \dfrac{24}{26} = \dfrac{12}{13}$

cos $y = \dfrac{Base}{Hypotenuse}$

$= \dfrac{BD}{AB} = \dfrac{32}{40} = \dfrac{4}{5}$

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

$= \dfrac{AD}{BD} = \dfrac{24}{32} = \dfrac{3}{4}$

Now,

$\dfrac{6}{\text{ cos x}} -\dfrac{5}{\text{ cos y}} + 8 {\text{ tan y}}\\[1em] = \dfrac{6}{\dfrac{12}{13}} -\dfrac{5}{\dfrac{4}{5}} + 8 \times \dfrac{3}{4}\\[1em] = \dfrac{6 \times 13}{12} -\dfrac{5 \times 5}{4} + 2 \times 3\\[1em] = \dfrac{13}{2} -\dfrac{25}{4} + 6\\[1em] = \dfrac{13 \times 2}{2 \times 2} -\dfrac{25}{4} + \dfrac{6 \times 4}{4}\\[1em] = \dfrac{26}{4} -\dfrac{25}{4} + \dfrac{24}{4}\\[1em] = \dfrac{26 - 25 + 24}{4}\\[1em] = \dfrac{25}{4}\\[1em] = 6\dfrac{1}{4}$

Hence, $\dfrac{6}{\text{ cos x}} -\dfrac{5}{\text{ cos y}} + 8 {\text{ tan y}} = 6\dfrac{1}{4}$