Problem 1 :
Prove :
(1 - cos2θ) ⋅ csc2θ = 1
Problem 2 :
Prove :
secθ ⋅ √(1 - sin2θ) = 1
Problem 3 :
Prove :
tanθ ⋅ sinθ + cosθ = secθ
Problem 4 :
Prove :
(1 - cosθ) ⋅ (1 + cosθ) ⋅ (1 + cot2θ) = 1
Problem 5 :
Prove :
cotθ + tanθ = secθ ⋅ cscθ
Problem 6 :
Prove :
cosθ/(1 - tanθ) + sinθ/(1 - cotθ) = sinθ + cosθ
Problem 7 :
Prove :
tan4θ + tan2θ = sec4θ - sec2θ
Problem 8 :
Prove :
√{(secθ – 1)/(secθ + 1)} = cscθ - cotθ
Problem 9 :
Prove :
(1 - sinA)/(1 + sinA) = (secA - tanA)2
Problem 10 :
Prove :
(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/cosθ
1. Answer :
(1 - cos2θ) ⋅ csc2θ = 1
Let A = (1 - cos2θ) ⋅ csc2θ and B = 1.
Because, sin2θ + cos2θ = 1, we have sin2θ = 1 - cos2θ
Then, we have
A = sin2θ ⋅ csc2θ
We know that csc2θ = 1/sin2θ
A = sin2θ ⋅ 1/sin2θ
A = 1
A = B Proved
2. Answer :
secθ ⋅ √(1 - sin2θ) = 1
Let A = secθ ⋅ √(1 - sin2θ) and B = 1.
Because sin2θ + cos2θ = 1, we have cos2θ = 1 - sin2θ.
A = secθ ⋅ √cos2θ
A = secθ ⋅ cosθ
A = secθ ⋅ 1/secθ
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 = (sinθ/cosθ) ⋅ sinθ + cosθ
A = (sin2θ/cosθ) + cosθ
A = (sin2θ + cos2θ)/cosθ
A = 1/cosθ
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)
Because, sin2θ + cos2θ = 1, we have sin2θ = 1 - cos2θ.
A = sin2θ ⋅ (1 + cot2θ)
A = sin2θ + sin2θ ⋅ cos2θ
A = sin2θ + sin2θ ⋅ (cos2θ/sin2θ)
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
A = cosθ/sinθ + sinθ/cosθ
A = (cos2θ + sin2θ)/(sinθ ⋅ cosθ)
(Because, cos2θ + sin2θ = 1)
A = 1/(sinθ ⋅ cosθ)
A = (1/cos θ) ⋅ (1/sin θ)
A = secθ ⋅ cscθ
A = B Proved
6. Answer :
cosθ/(1 - tanθ) + sinθ/(1 - cotθ) = sinθ + cosθ
Let A = cosθ/(1 - tanθ) + sinθ/(1
- cotθ) and
B = sinθ + cosθ.
A = cosθ/{1 - (sinθ/cos θ)} + sinθ/{1
- (cosθ/sinθ)}
A = cosθ/{(cosθ - sinθ)/cosθ} +
sinθ/{(sinθ - cosθ/sinθ)}
A = cos2θ/(cosθ - sinθ) + sin2θ/(sinθ - cosθ)
A = cos2θ/(cosθ - sinθ) - sin2θ/(cosθ - sinθ)
A = (cos2θ - sin2θ)/(cosθ - sinθ)
A = [(cosθ + sinθ)(cosθ - sinθ)]/(cosθ - sinθ)
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)
[Because tan2θ = sec2θ – 1]
A = (sec2θ - 1) ⋅ sec2θ
[Because, tan2θ + 1 = sec2θ]
A = sec4θ - sec2θ
A = B Proved
8. Answer :
√{(secθ – 1)/(secθ + 1)} = cscθ - cotθ
Let A = √{(secθ
– 1)/(secθ + 1)} and B = cosecθ - cotθ.
A = √{(secθ – 1)/(secθ + 1)}
A = √[{(secθ - 1) (secθ - 1)}/{(secθ + 1) (secθ - 1)}]
[Multiplying numerator and denominator by (secθ - l) inside radical sign]
A = √{(secθ - 1)2/(sec2θ - 1)}
A = √{(secθ -1)2/tan2θ}
[Because, sec2 θ = 1 + tan2 θ ⇒ sec2θ - 1 = tan2θ]
A = (secθ – 1)/tanθ
A = (secθ/tanθ) – (1/tanθ)
A = {(1/cosθ)/(sinθ/cos θ)} - cotθ
A = {(1/cosθ) ⋅ (cosθ/sinθ)} - cotθ
A = (1/sinθ) - cotθ
A = cosecθ - cotθ
A = B Proved
9. Answer :
(1 - sinA)/(1 + sinA) = (secA - tanA)2
Let A = (1 - sinA)/(1
+ sinA) and B = (secA - tanA)2.
A = (1 - sinA)/(1 + sinA)
A = (1 - sinA)2/(1 - sinA)(1 + sinA)
[Multiply both numerator and denominator by (1 - sin A)
A = (1 - sinA)2/(1 - sin2A)
A = (1 - sinA)2/(cos2A)
[Because, sin2θ + cos2θ = 1 ⇒ cos2θ = 1 - sin2θ]
A = {(1 - sinA)/cosA}2
A = (1/cosA - sinA/cosA)2
A = (secA – tanA)2
A = B Proved
10. Answer :
(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/cosθ
Let A = (tanθ + secθ
- 1)/(tanθ - secθ + 1) and
B = (1 + sinθ)/cosθ.
A = (tanθ + secθ - 1)/(tanθ - secθ + 1)
A = [(tanθ + secθ) - (sec2θ - tan2θ)]/(tanθ - secθ + 1)
[Because sec2θ - tan2θ = 1]
A = {(tanθ + secθ) - (secθ + tanθ)(secθ - tanθ)}/(tanθ - secθ + 1)
A = {(tanθ + secθ) (1 - secθ +
tanθ)}/(tanθ - secθ + 1)
A = {(tanθ + secθ) (tanθ - secθ
+ 1)}/(tanθ - secθ + 1)
A = tanθ + secθ
A = (sinθ/cosθ) + (1/cosθ)
A = (sinθ + 1)/cosθ
A = (1 + sinθ)/cosθ
A = B Proved
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