Prove each of the following.
Problem 1 :
(1 - cos2θ) ⋅ csc2θ = 1
Problem 2 :
Problem 3 :
tanθ ⋅ sinθ + cosθ = secθ
Problem 4 :
(1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) = 1
Problem 5 :
cotθ + tanθ = secθ ⋅ cscθ
Problem 6 :
Problem 7 :
tan4θ + tan2θ = sec4θ - sec2θ
Problem 8 :
Problem 9 :
Problem 10 :
1. Answer :
(1 - cos2θ) ⋅ csc2θ = 1
Let A = (1 - cos2θ) ⋅ csc2θ and B = 1.
Since sin2θ + cos2θ = 1, we have sin2θ = 1 - cos2θ.
Then, we have
A = sin2θ ⋅ csc2θ
A = 1
A = B Proved
2. Answer :
Since sin2θ + cos2θ = 1, we have cos2θ = 1 - sin2θ.
A = 1
A = B Proved
3. Answer :
tanθ ⋅ sinθ + cosθ = secθ
Let A = tanθ ⋅ sin θ + cos θ and B = sec θ.
A = tanθ ⋅ sin θ + cos θ
A = secθ
A = B Proved
4. Answer :
(1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) = 1
Let A = (1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) and B = 1.
A = (1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ)
A = (1 - cos2θ) ⋅ (1 + cot2θ)
Since, sin2θ + cos2θ = 1, we have sin2θ = 1 - cos2θ.
A = sin2θ ⋅ (1 + cot2θ)
A = sin2θ + cos2θ
A = 1
A = B Proved
5. Answer :
cotθ + tanθ = secθ ⋅ cscθ
Let A = cotθ + tanθ and B = secθ ⋅ cscθ.
A = cotθ + tanθ
Since cos2θ + sin2θ = 1, we have
A = secθ ⋅ cscθ
A = B Proved
6. Answer :
A = cosθ + sinθ
A = B Proved
7. Answer :
tan4θ + tan2θ = sec4θ - sec2θ
Let A = tan2θ + tan2θ and B = sec2θ - sec2θ.
A = tan2θ (tan2θ + 1)
A = (sec2θ - 1)(tan2θ + 1)
Since tan2θ = sec2θ – 1, we have
A = (sec2θ - 1) ⋅ sec2θ
Since tan2θ + 1 = sec2θ, we have
A = sec4θ - sec2θ
A = B Proved
8. Answer :
Since sec2θ = 1 + tan2θ, we have sec2θ - 1 = tan2θ.
A = cscθ - cotθ
A = B Proved
9. Answer :
A = (secA – tanA)2
A = B Proved
10. Answer :
Since sec2θ - tan2θ = 1, we have
A = B Proved
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