Solution:

(i) Required to evaluate:
, given = cot θ = 7/8

Taking the numerator, we have

(1+sin θ)(1–sin θ) = 1 – sin² θ [Since, (a+b)(a-b) = a² – b²]

Similarly,

(1+cos θ)(1–cos θ) = 1 – cos² θ

We know that,

sin² θ + cos² θ = 1

⇒ 1 – cos² θ = sin² θ

And,

1 – sin² θ = cos² θ

Thus,

(1+sin θ)(1 –sin θ) = 1 – sin² θ = cos² θ

(1+cos θ)(1–cos θ) = 1 – cos² θ = sin² θ

⇒

= cos² θ/ sin² θ

= (cos θ/sin θ)²

And, we know that (cos θ/sin θ) = cot θ

⇒

= (cot θ)²

= (7/8)²

= 49/ 64

(ii) Given,

cot θ = 7/8

So, by squaring on both sides we get

(cot θ)² = (7/8)²

∴ cot θ² = 49/64