Let ABC be a right-angled triangle. Given, in A = 4/5 Or sin A = BC/AC = (4/5) By using Pythagoras theorem, BA ² = AC ² – BC ² = 25 – 16= 9 So, BA = 3units Now, Cosec A × cos A + sec A × tan A = (AC/BC) ×(AB/AC) + (AC/AB) × (BC/AB) = (5/4) × (3/5) +(5/3) × (4/3) = (3/4) + (20/9) =107/36
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