Question

sin x =a^2-b^2,find other trigonometric ratios

Answer 1

ΔABC, Let ∠ABC be θ Sin θ = (a2 - b2) / (a2 + b2) ⇒ AB = (a2 - b2) ⇒ AC = (a2 + b2) ⇒ BC = √[(a2 + b2)2 - (a2 - b2)2]   [According to Pythagoras theorem] ⇒ BC = √(4a2b2) ⇒ BC = 2ab Cos θ = 2ab / (a2 + b2) Tan θ = (a2 - b2) / 2ab. Cosec θ, Sec θ and Cot θ are the reciprocals of Sin θ, Cos θ, Tan θ respectively.