AOB is a diameter of a circle with centre O and C is any point on the circle, joining A.C; B,C; and O, C let us show that

(i) tan∠ABC = cot ∠ACO

(ii) sin²∠BCO + sin²∠ACO = 1

(iii) cosec²∠CAB - 1 = tan²∠ABC

(i) tan∠ABC = cot ∠ACO

⇒ ∠ AOC = 90° = ∠ BOC

⇒ ∠ CAO = 60° and

∠ ACO = 30°

⇒ tan∠ ABC = tan(90 - ∠ ACO)

[tan(90 - θ) = cotθ ]

= cot∠ ACO

(ii) sin²∠BCO + sin²∠ACO = 1

⇒ sin²∠BCO + sin²∠ACO = 1

⇒ sin²∠BCO + sin²∠ACO

⇒ sin²∠BCO + sin²(90 - ∠ BCO)

⇒ sin²∠BCO + cos²∠BCO

= 1

[since, sin²θ + cos²θ = 1]

(iii) cosec²∠CAB - 1 = tan²∠ABC

⇒ cosec²∠CAB - 1

⇒ cosec(90 - ∠CAB) = sec∠CAB

⇒ ∠ CAB = ∠ ABC

⇒ sec²∠CAB - 1 = tan²∠ ABC

⇒ sec²∠ ABC – 1 = tan²∠ ABC

⇒ tan²∠ ABC = tan²∠ ABC