From the following figure, find :

(i) y

(ii) sin x°

(iii) (sec x° - tan x°)(sec x° + tan x°)

(i) In Δ ABC,

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

⇒ 2² = y² + 1²

⇒ 4 = y² + 1

⇒ y² = 4 - 1

⇒ y² = 3

⇒ y = $\sqrt{3}$

Hence, the value of y = $\sqrt{3}$.

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

$= \dfrac{AB}{AC}\\[1em] = \dfrac{\sqrt{3}}{2}$

Hence, sin $x° = \dfrac{\sqrt{3}}{2}$.

(iii) (sec x° - tan x°)(sec x° + tan x°)

sec $x° = \dfrac{Hypotenuse}{Base}$

$= \dfrac{AC}{CB}\\[1em] = \dfrac{2}{1} = 2$

tan $x° = \dfrac{Perpendicular}{Base}$

$= \dfrac{AB}{CB}\\[1em] = \dfrac{\sqrt{3}}{1} = 3$

Now, (sec x° - tan x°)(sec x° + tan x°)

$= (2 - \sqrt{3})(2 + \sqrt{3}) \\[1em] = (2)^2 - (\sqrt{3})^2 \\[1em] = 4 - 3 \\[1em] = 1$

Hence, (sec x° - tan x°)(sec x° + tan x°) = 1.