Solution:

(i) Option (B) is correct.

Proof:

When you take 9 outside, it becomes

9 sec²A – 9 tan²A

= 9 (sec²A – tan²A)

= 9×1 = 9             (∵ sec2 A – tan2 A = 1)

As a result, 9 sec²A – 9 tan²A = 9

(ii) Option (C) is correct

Proof:

(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

We even know that, tan θ = sin θ/cos θ

sec θ = 1/ cos θ

cot θ = cos θ/sin θ

cosec θ = 1/sin θ

Substituting the above values into the problem gives:

= (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ)

Simplifying the provided equation as,

= (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

= (cos θ+sin θ)²-1²/(cos θ sin θ)

= (cos²θ + sin²θ + 2cos θ sin θ -1)/(cos θ sin θ)

= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos²θ + sin²θ = 1)

As a result,