Q. 10

# AOB is a diameter

(i) tanABC = cot ACO

AOC = 90° = BOC

CAO = 60° and

ACO = 30°

tan ABC = tan(90 - ACO)

[tan(90 - θ) = cotθ ]

= cot ACO

(ii) sin2BCO + sin2ACO = 1

sin2BCO + sin2ACO = 1

sin2BCO + sin2ACO

sin2BCO + sin2(90 - BCO)

sin2BCO + cos2BCO

= 1

[since, sin2θ + cos2θ = 1]

(iii) cosec2CAB - 1 = tan2ABC

cosec2CAB - 1

cosec(90 - CAB) = secCAB

CAB = ABC

sec2CAB - 1 = tan2 ABC

sec2 ABC – 1 = tan2 ABC

tan2 ABC = tan2 ABC

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