Given that,

tan A = cot B

tan A = tan (90° − B)

A = 90° − B

A + B = 90°

Given that,

sec 4A = cosec (A − 20°)

cosec (90° − 4A) = cosec (A − 20°)

90° − 4A= A− 20°

110° = 5A

A = 22°

In a triangle, sum of all the interior angles

A + B + C = 180°

⇒ B + C = 180° - A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

⇒ sin (B+C)/2 = sin (90°-A/2)

⇒ sin (B+C)/2 = cos A/2

sin 67° + cos 75°
= sin (90° - 23°) + cos (90° - 15°)
= cos 23° + sin 15°

cosec²A - cot²A = 1
⇒ cosec²A = 1 + cot²A
⇒ 1/sin²A = 1 + cot²A
⇒sin²A = 1/(1+cot²A)

Now,
sin²A = 1/(1+cot²A)
⇒ 1 - cos²A = 1/(1+cot²A)
⇒cos²A = 1 - 1/(1+cot²A)
⇒cos²A = (1-1+cot²A)/(1+cot²A)
⇒ 1/sec²A = cot²A/(1+cot²A)
⇒ secA = (1+cot²A)/cot²A

also,
tan A = sin A/cos A and cot A = cos A/sin A
⇒ tan A = 1/cot A

We know that sec²A−tan²A=1

⇒tan²A=sec²A−1

⇒tan²A=sec²A−1

⇒tanA = √ sec²A−1−−−−−−−−    (1)

We know that cos A = 1 / secA      (2)

We know that sin²A+cos²A=1

⇒sin2A=1−cos2A

Putting (2) in the above equation we get

 

We know that  cosecA=1/ sinA

Putting (3) in the above equation, we get

cosecA=secA / √ sec2A−1        (4) 

 

We know that  cotA=1 /tanA

Putting (1) in the above equation, we get

cotA=1 / √sec²A−1                    (5)

 (1), (2), (3), (4) and (5) shows all the other trigonometric ratios of ∠A in terms of sec A.

(i) (sin²63° + sin²27°)/(cos²17° + cos²73°)
   = [sin²(90°-27°) + sin²27°]/[cos²(90°-73°) + cos²73°)]
   = (cos²27° + sin²27°)/(sin²27° + cos²73°)
   = 1/1 =1                       (��� sin²A + cos²A = 1)

(ii) sin 25° cos 65° + cos 25° sin 65°
   = sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
   = cos 65° cos 65° + sin 65° sin 65°
   = cos²65° + sin²65° = 1

(i) (B) is correct.

9 sec²A - 9 tan²A

= 9 (sec²A - tan²A)
= 9×1 = 9             (��� sec2 A - tan2 A = 1)

(ii) (C) is correct

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

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

= (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)

= (2cos θ sin θ)/(cos θ sin θ) = 2

(iii) (D) is correct.

(secA + tanA) (1 - sinA)

= (1/cos A + sin A/cos A) (1 - sinA)

= (1+sin A/cos A) (1 - sinA)

= (1 - sin²A)/cos A

= cos²A/cos A = cos A

(iv) (D) is correct.

1+tan²A/1+cot²A 

= (1+1/cot²A)/1+cot²A

= (cot²A+1/cot²A)×(1/1+cot²A)

= 1/cot²A = tan²A