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