From the figure (2) given below, find the values of :

(i) sin A

(ii) cos A

(iii) sin² A + cos² A

(iv) sec² A - tan² A

In right-angled triangle ABC,

By Pythagoras theorem, we get

⇒ AB² = AC² + BC²

⇒ AB² = (12)² + (5)²

⇒ AB² = 144 + 25

⇒ AB² = 169

⇒ AB = $\sqrt{169}$ = 13

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

= $\dfrac{BC}{AB} = \dfrac{5}{13}$.

Hence, sin A = $\dfrac{5}{13}$.

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

= $\dfrac{AC}{AB} = \dfrac{12}{13}$.

Hence, cos A = $\dfrac{12}{13}$.

(iii) Substituting values of sin A and cos A in sin² A + cos² A :

⇒ sin² A + cos² A = $\Big(\dfrac{5}{13}\Big)^2 + \Big(\dfrac{12}{13}\Big)^2$

= $\dfrac{25}{169} + \dfrac{144}{169}$

= $\dfrac{169}{169}$ = 1.

Hence, sin² A + cos² A = 1.

(iv) sec² A - tan² A = $\Big(\dfrac{AB}{AC}\Big)^2 - \Big(\dfrac{BC}{AC}\Big)^2$

= $\Big(\dfrac{13}{12}\Big)^2 - \Big(\dfrac{5}{12}\Big)^2$

= $\dfrac{169}{144} - \dfrac{25}{144}$

= $\dfrac{144}{144}$ = 1.

Hence, sec² A - tan² A = 1.