In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = $\dfrac{4}{5}$ and tan C = 1.

(i) Calculate the lengths of AD, AB, DC and AC.

(ii) Show that $\text{tan}^2 \text{ B} - \dfrac{1}{\text{cos}^2 \text{ B}}$ = -1.

(i) From figure,

tan C = $\dfrac{\text{Perpendicular}}{\text{Base}}$ = $\dfrac{AD}{DC}$

⇒ 1 = $\dfrac{AD}{DC}$

⇒ AD = DC.

sin B = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$ = $\dfrac{AD}{AB}$

⇒ $\dfrac{4}{5}$ = $\dfrac{AD}{AB}$

Let AD = 4k and AB = 5k.

In right angle triangle ABD,

⇒ AB² = AD² + BD²

⇒ (5k)² = (4k)² + 15²

⇒ 25k² = 16k² + 225

⇒ 9k² = 225

⇒ k² = $\dfrac{225}{9}$ = 25

⇒ k = $\sqrt{25}$ = 5.

AD = 4k = 4 × 5 = 20

AB = 5k = 5 × 5 = 25.

DC = AD = 20.

In right angle triangle ADC,

⇒ AC² = AD² + DC²

⇒ AC² = 20² + 20²

⇒ AC² = 400 +400

⇒ AC² = 800

⇒ AC = $\sqrt{800} = 20\sqrt{2}$.

Hence, AD = 20, AB = 25, DC = 20 and AC = $20\sqrt{2}$.

(ii) Calculating tan B, we get :

$\text{tan B} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] = \dfrac{AD}{BD} \\[1em] = \dfrac{20}{15} = \dfrac{4}{3}.$

Calculating cos B, we get :

$\text{cos B} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BD}{AB} \\[1em] = \dfrac{15}{25} = \dfrac{3}{5}.$

Substituting value of tan B and cos B in L.H.S. of the equation, $\text{tan}^2 \text{ B} - \dfrac{1}{\text{cos}^2 \text{ B}}$ = -1, we get :

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

Hence, proved that tan² B - $\dfrac{1}{\text{cos}^2 B}$ = -1.