Q. 10

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) tanABC = cot ACO

(ii) sin2BCO + sin2ACO = 1

(iii) cosec2CAB - 1 = tan2ABC

Answer :

geogebra-export.png


(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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