If cot θ = $\dfrac{7}{8}$, evaluate :

(i) $\dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}}$

(ii) cot² θ

Let us draw a right angle triangle ABC, with ∠A = θ.

We know that,

cot θ = $\dfrac{\text{Side adjacent to ∠θ}}{\text{Side opposite to ∠θ}}$

Substituting values we get :

$\dfrac{7}{8} = \dfrac{AB}{BC}$

Let AB = 7k and BC = 8k.

In right angle triangle ABC,

⇒ AC² = AB² + BC ²

⇒ AC² = (7k)² + (8k)²

⇒ AC² = 49k² + 64k²

⇒ AC² = 113k²

⇒ AC = $\sqrt{113}k$.

(i) We know that,

sin θ = $\dfrac{\text{Side opposite to ∠θ}}{\text{Hypotenuse}} = \dfrac{BC}{AC} = \dfrac{8k}{\sqrt{113}k} = \dfrac{8}{\sqrt{113}}$.

cos θ = $\dfrac{\text{Side adjacent to ∠θ}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{7k}{\sqrt{113}k} = \dfrac{7}{\sqrt{113}}$.

Substituting values of sin θ and cos θ in $\dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}}$, we get :

$\Rightarrow \dfrac{\Big(1 + \dfrac{8}{\sqrt{113}}\Big)\Big(1 - \dfrac{8}{\sqrt{113}}\Big)}{\Big(1 + \dfrac{7}{\sqrt{113}}\Big)\Big(1 - \dfrac{7}{\sqrt{113}}\Big)} \\[1em] \Rightarrow \dfrac{(1)^2 - \Big(\dfrac{8}{\sqrt{113}}\Big)^2}{(1)^2 - \Big(\dfrac{7}{\sqrt{113}}\Big)^2} \\[1em] \Rightarrow \dfrac{1 - \dfrac{64}{113}}{1 - \dfrac{49}{113}} \\[1em] \Rightarrow \dfrac{\dfrac{113 - 64}{113}}{\dfrac{113 - 49}{113}} \\[1em] \Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}} \\[1em] \Rightarrow \dfrac{49}{64}.$

Hence, $\dfrac{\text{(1 + sin θ)(1 - sin θ)}}{\text{(1 + cos θ)(1 - cos θ)}} = \dfrac{49}{64}$.

(ii) Given,

⇒ cot θ = $\dfrac{7}{8}$

⇒ cot² θ = $\Big(\dfrac{7}{8}\Big)^2 = \dfrac{49}{64}$.

Hence, cot² θ = $\dfrac{49}{64}$.