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Question

# If secθ+tanθ=p , prove that (i) secθ=12(p+1p) (ii) tanθ=12(p−1p) (iii) sinθ=p2−1p2+1

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Solution

## 1 secθ+tanθ=p-------------(1) we know, sec²θ - tan²θ = 1 (secθ - tanθ)(secθ + tanθ) = 1 put , secθ + tanθ = P (secθ - tanθ) × P = 1 (secθ - tanθ) = 1P ---------(2) add equations (1) and (2) 2secθ = P + 1P secθ=12(p+1p) 2- Secθ + tanθ = P -------------(1) we know, sec²θ - tan²θ = 1 (secθ - tanθ)(secθ + tanθ) = 1 put , secθ + tanθ = P (secθ - tanθ) × P = 1 (secθ - tanθ) = 1P ---------(2) add equations (1) and (2) 2secθ = P + 1P secθ= (P²+1)2P cosθ = 1secθ = 2P(P²+1) we know, cosθ =basehypotenuse so, perpendicular = √(h²−b²) = √(P²+1)²−(2P)² =√(P²−1)² =(P² -1) sinθ =perpendicularhypotenuse = (p² -1)/(P² +1) hence, sinθ = (P² -1)/(P² + 1) sotanθ =sinθcosθ (tanθ=12(p−1p) 3- Secθ + tanθ = P -------------(1) we know, sec²θ - tan²θ = 1 (secθ - tanθ)(secθ + tanθ) = 1 put , secθ + tanθ = P (secθ - tanθ) × P = 1 (secθ - tanθ) =1P---------(2) add equations (1) and (2) 2secθ = P + 1P secθ = ( P² + 1)/2P cosθ = 1/secθ =( 2P)/(P² + 1) we know, cosθ = base/hypotenuse so, perpendicular = √(h²−b²) = √(P²+1)²−(2P)² =√(P²−1)²) =(P² -1) sinθ =perpendicular/hypotenuse = (p² -1)/(P² +1) hence, sinθ=p2−1p2+1

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