TRIGONOMETRY QUESTIONS AND ANSWERS

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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