Question 1 :
If tan2 θ = 1 − k2, show that
sec θ + tan3 θ cosec θ = (2 - k2)3/2
Also, find the values of k for which this result holds.
Answer :
sec θ + tan3 θ cosec θ
= (1/cos θ) + (sin θ/cos θ)3 (1/sin θ)
= (1/cos θ) + (sin3θ/cos3 θ) (1/sin θ)
= (1/cos θ) + (sin2θ/cos3 θ)
= (cos2 θ + sin2θ)/cos3 θ
= 1/cos3 θ
= sec3θ
= (sec2θ) sec θ ----(1)
Given that :
tan2 θ = 1 − k2
Add 1 on both sides
1 + tan2 θ = 1 + 1 − k2
sec2θ = 2 − k2 ==> secθ = √(2 − k2)
By applying the above values in the first equation, we get
= (2 − k2) √(2 − k2)
= (2 − k2)3/2
The value of k is [-1, 1]
Question 2 :
If sec θ + tanθ = p, obtain the values of sec θ, tan θ and sin θ in terms of p.
Answer :
Given that :
sec θ + tanθ = p ------(1)
(sec θ + tanθ) ⋅ (sec θ - tanθ)/(sec θ - tanθ) = p
(sec2 θ - tan2 θ)/(sec θ - tanθ) = p
1/(sec θ - tanθ) = p
1/p = sec θ - tanθ ---(2)
(1) + (2)
By adding first and second equation, we get the value of sec θ
sec θ + tanθ + sec θ - tanθ = p + (1/p)
2 sec θ = (p2 + 1)/p
Hence the value of sec θ = (p2 + 1)/2p
(1) - (2)
By subtracting second equation from the first equation, we get the value of tan θ
sec θ + tanθ - (sec θ - tanθ) = p - (1/p)
sec θ + tanθ - sec θ + tanθ = p - (1/p)
2 tan θ = (p2 - 1)/p
Hence the value of tan θ = (p2 - 1)/2p
By dividing tan θ by sec θ, we get the value of sin θ
tan θ/sec θ = [(p2 - 1)/2p] / [(p2 + 1)/2p]
= [(p2 - 1)/2p] / [2p/(p2 + 1)]
= (p2 - 1)/(p2 + 1)
Hence the value of sin θ is (p2 - 1)/(p2 + 1).
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