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Question

If secθ+tanθ=p , prove that

(i) secθ=12(p+1p)

(ii) tanθ=12(p1p)

(iii) sinθ=p21p2+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 = (h²b²)
= (P²+1)²(2P)²
=(P²1)²
=(P² -1)

sinθ =perpendicularhypotenuse
= (p² -1)/(P² +1)

hence,
sinθ = (P² -1)/(P² + 1)
sotanθ =sinθcosθ
(tanθ=12(p1p)
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 = (h²b²)
= (P²+1)²(2P)²
=(P²1)²)
=(P² -1)

sinθ =perpendicular/hypotenuse
= (p² -1)/(P² +1)

hence,
sinθ=p21p2+1


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