If 4 cos² x° - 1 = 0 and 0 ≤ x° ≤ 90°, find:

(i) x°

(ii) sin² x° + cos² x°

(iii) $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°}$

(i) 4 cos² x° - 1 = 0

⇒ 4 cos² x° = 1

⇒ cos² x° = $\dfrac{1}{4}$

⇒ cos x° = $\sqrt{\dfrac{1}{4}}$

⇒ cos x° = $\dfrac{1}{2}$

⇒ cos x° = cos 60°

Hence, x° = 60°.

(ii) sin² x° + cos² x°

⇒ sin² 60° + cos² 60°

$= \Big(\dfrac{\sqrt3}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{3}{4} + \dfrac{1}{4} \\[1em] = \dfrac{3 + 1}{4}\\[1em] = \dfrac{4}{4}\\[1em] = 1$

Hence, sin² x° + cos² x° = 1.

(iii) $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°}$

$= \dfrac{1}{\text{cos}^2 \text{ 60°}} - \text{tan}^2 \text{ 60°}\\[1em] = \dfrac{1}{\Big(\dfrac{1}{2}\Big)^2} - (\sqrt3)^2\\[1em] = \dfrac{4}{1} - 3\\[1em] = 4 - 3\\[1em] = 1$

Hence, $\dfrac{1}{\text{cos}^2 \text{ x°}} - \text{tan}^2 \text{ x°} = 1$.