(i) If cos A = $\dfrac{9}{41}$; find the value of :

$\dfrac{1}{\text{sin}^2 A} - \text{cot}^2 A$

(ii) If (2cos 2A - 1) (tan3A - 1) = 0; find all possible values of angle A.

(i) Given: cos A = $\dfrac{9}{41}$

⇒ $\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{9}{41}$

Let base be 9a and hypotenuse be 41a.

Using Pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

⇒ (41a)² = (9a)² + Perpendicular²

⇒ 1681a² = 81a² + Perpendicular²

⇒ Perpendicular² = 1681a² - 81a²

⇒ Perpendicular² = 1600a²

⇒ Perpendicular = $\sqrt{1600a^2}$

⇒ Perpendicular = 40a

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

= $\dfrac{40a}{41a}$

= $\dfrac{40}{41}$

⇒ (sin A)² = $\Big(\dfrac{40}{41}\Big)^2$

⇒ sin² A = $\dfrac{1600}{1681}$

And, cot A = $\dfrac{\text{Base}}{\text{Perpendicular}}$

= $\dfrac{9a}{40a}$

= $\dfrac{9}{40}$

⇒ (cot A)² = $\Big(\dfrac{9}{40}\Big)^2$

⇒ cot² A = $\dfrac{81}{1600}$

Now, the value of

$= \dfrac{1}{\text{sin}^2 A} - \text{cot}^2 A\\[1em] = \dfrac{1}{\dfrac{1600}{1681}} - \Big(\dfrac{81}{1600}\Big)\\[1em] = \Big(\dfrac{1681}{1600}\Big) - \Big(\dfrac{81}{1600}\Big)\\[1em] = \dfrac{1681 - 81}{1600}\\[1em] = \dfrac{1600}{1600}\\[1em] = 1$

Hence, the value of $\dfrac{1}{\text{sin}^2 A} - \text{cot}^2 A = 1$.

(ii) Given: (2cos 2A - 1) (tan3A - 1) = 0

⇒ (2cos 2A - 1) = 0 or (tan3A - 1) = 0

⇒ 2cos 2A = 1 or tan3A = 1

⇒ cos 2A = $\dfrac{1}{2}$ or tan3A = tan 45°

⇒ cos 2A = cos 60° or tan3A = tan 45°

⇒ 2A = 60° or 3A = 45°

⇒ A = $\dfrac{60°}{2} \text{ or } A = \dfrac{45°}{3}$

⇒ A = 30° or A = 15°

Hence, the value of A = 30° or 15°.