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.

sec θ + tan3 θ cosec θ

=  (1/cos θ) + (sin θ/cos θ)3 (1/sin θ)

=  (1/cos θ) + (sin3θ/cosθ) (1/sin θ)

=  (1/cos θ) + (sin2θ/cosθ)

=  (cos2 θ + sin2θ)/cosθ

=  1/cosθ

=  sec3θ

=  (sec2θ) sec θ  ----(1)

Given that :

tan2 θ = 1 − k2

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.

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)

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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