From the following figure, find the values of :

(i) sin B

(ii) tan C

(iii) sec² B – tan² B

(iv) sin² C + cos² C

In Δ ABD,

⇒ AB² = BD² + DA² (∵ AB is hypotenuse)

⇒ 13² = 5² + DA²

⇒ 169 = 25 + DA²

⇒ DA² = 169 - 25

⇒ DA² = 144

⇒ DA = $\sqrt{144}$

⇒ DA = 12

In Δ ADC,

⇒ AC² = AD² + DC² (∵ AB is hypotenuse)

⇒ AC² = 12² + 16²

⇒ AC² = 144 + 256

⇒ AC² = 400

⇒ AC = $\sqrt{400}$

⇒ AC = 20

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

$= \dfrac{AD}{AB}\\[1em] = \dfrac{12}{13}$

Hence, sin $B = \dfrac{12}{13}$.

(ii) tan $C = \dfrac{Perpendicular}{Base}$

$= \dfrac{AD}{DC}\\[1em] = \dfrac{12}{16}\\[1em] = \dfrac{3}{4}$

Hence, tan $C = \dfrac{3}{4}$.

(iii) sec² B – tan² B

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

$= \dfrac{AB}{BD}\\[1em] = \dfrac{13}{5}$

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

$= \dfrac{AD}{BD}\\[1em] = \dfrac{12}{5}$

sec² B – tan² B

$= \Big(\dfrac{13}{5}\Big)^2 - \Big(\dfrac{12}{5}\Big)^2\\[1em] = \Big(\dfrac{169}{25}\Big) - \Big(\dfrac{144}{25}\Big)\\[1em] = \Big(\dfrac{169 - 144}{25}\Big)\\[1em] = \Big(\dfrac{25}{25}\Big)\\[1em] = 1$

Hence, sec² B – tan² B = 1.

(iv) sin² C + cos² C

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

$= \dfrac{AD}{AC}\\[1em] = \dfrac{12}{20}\\[1em] = \dfrac{3}{5}$

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

$= \dfrac{DC}{AC}\\[1em] = \dfrac{16}{20}\\[1em] = \dfrac{4}{5}$

Now,

sin² C + cos² C

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

Hence, sin² C + cos² C = 1.