From the following figure, find the values of :

(i) sin A

(ii) sec A

(iii) cos² A + sin² A

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = a² + a²

⇒ AC² = 2a²

⇒ AC = $\sqrt{2a^2}$

⇒ AC = $a \sqrt{2}$

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

$= \dfrac{CB}{AC}\\[1em] = \dfrac{a}{a \sqrt{2}}\\[1em] = \dfrac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{a \sqrt{2}}{a \times 2}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \dfrac{\sqrt{2}}{2}\\[1em]$

Hence, sin $A = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$.

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

$= \dfrac{AC}{AB}\\[1em] = \dfrac{a \sqrt{2}}{a}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a}}\\[1em] = \sqrt{2}$

Hence, sec $A = \sqrt{2}$.

(iii) cos² A + sin² A

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

$= \dfrac{AB}{AC}\\[1em] = \dfrac{a}{a \sqrt{2}}\\[1em] = \dfrac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{a \sqrt{2}}{a \times 2}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \dfrac{\sqrt{2}}{2}\\[1em]$

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

$= \dfrac{CB}{AC}\\[1em] = \dfrac{a}{a \sqrt{2}}\\[1em] = \dfrac{a \times \sqrt{2}}{a \sqrt{2} \times \sqrt{2}}\\[1em] = \dfrac{a \sqrt{2}}{a \times 2}\\[1em] = \dfrac{\cancel{a} \sqrt{2}}{\cancel{a} \times 2}\\[1em] = \dfrac{\sqrt{2}}{2}\\[1em]$

Now, cos² A + sin² A

$= \Big(\dfrac{\sqrt{2}}{2}\Big)^2 + \Big(\dfrac{\sqrt{2}}{2}\Big)^2\\[1em] = \dfrac{2}{4} + \dfrac{2}{4}\\[1em] = \dfrac{2 + 2}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

Hence, cos² A + sin² A = 1.