Trigonometry
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If cos θ + cos θ = 4 , then the value of θ (0° < θ < 90 ° ) is 1 - sin θ 1 + sin θ
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cos θ + cos θ = 4 1 - sin θ 1 + sin θ ⇒ cos θ(1 + sin θ) + cos θ(1 - sin θ) = 4 (1 - sin θ)(1 + sin θ) ⇒ cos θ + cos θ.sin θ + cos θ - cos θ.sin θ = 4 (1 - sin2 θ) ⇒ 2cos θ = 4 cos2 θ ⇒ 1 = 2 cos θ ⇒ cos θ = 1 = cos 60° ⇒ θ = 60° 2 Correct Option: A
cos θ + cos θ = 4 1 - sin θ 1 + sin θ ⇒ cos θ(1 + sin θ) + cos θ(1 - sin θ) = 4 (1 - sin θ)(1 + sin θ) ⇒ cos θ + cos θ.sin θ + cos θ - cos θ.sin θ = 4 (1 - sin2 θ) ⇒ 2cos θ = 4 cos2 θ ⇒ 1 = 2 cos θ ⇒ cos θ = 1 = cos 60° ⇒ θ = 60° 2
- If x2 = sin230° + 4 cot245° – sec260°, then the value of x (x > 0) is
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x2 = sin230° + 4 cot245° – sec260°
= 
1 
2 + 4(1)2 – (2)2 2 = 1 + 4 – 4 = 1 4 4 ∴ x = 1 2 Correct Option: D
x2 = sin230° + 4 cot245° – sec260°
= 1 2 + 4(1)2 – (2)2 2 = 1 + 4 – 4 = 1 4 4 ∴ x = 1 2
- If 7sin2θ + 3cos2θ = 4 then the value of secθ + cosecθ is
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7sin2θ + 3cos2θ = 4
⇒ 4 sin2θ + 3 sin2θ + 3 cos2θ = 4
⇒ 4 sin2θ + 3(sin2θ + cos2θ) = 4
⇒ 4 sin2θ = 4 – 3
[∵ sin2θ + cos2θ = 1]
⇒ 4 sin2θ = 1⇒ sin2θ = 1 4 ⇒ sinθ = 1 = sin 30° 2
⇒ θ = 30°
∴ secθ + cosecθ
= sec 30° + cosec 30°= 2 + 2 √3 Correct Option: B
7sin2θ + 3cos2θ = 4
⇒ 4 sin2θ + 3 sin2θ + 3 cos2θ = 4
⇒ 4 sin2θ + 3(sin2θ + cos2θ) = 4
⇒ 4 sin2θ = 4 – 3
[∵ sin2θ + cos2θ = 1]
⇒ 4 sin2θ = 1⇒ sin2θ = 1 4 ⇒ sinθ = 1 = sin 30° 2
⇒ θ = 30°
∴ secθ + cosecθ
= sec 30° + cosec 30°= 2 + 2 √3
- If tanθ + cotθ = 5, then the value of tan2θ + cot2θ is
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tanθ + cotθ = 5
On squaring both sides,
(tanθ + cotθ)2 = 25
⇒ tan2θ + cot2θ + 2tanθ. cotθ = 25
⇒ tan2θ + cot2θ + 2 = 25
⇒ tan2θ + cot2θ = 25 – 2 = 23Correct Option: C
tanθ + cotθ = 5
On squaring both sides,
(tanθ + cotθ)2 = 25
⇒ tan2θ + cot2θ + 2tanθ. cotθ = 25
⇒ tan2θ + cot2θ + 2 = 25
⇒ tan2θ + cot2θ = 25 – 2 = 23
- If θ be positive acute angle and 5 cosθ + 12 sinθ = 13, then the value of cosθ is
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5 cosθ + 12 sinθ = 13
⇒ 5 cosθ – 13 = 12 sinθ
On squaring both sides,
25 cos2θ + 169 – 130 cosθ
= 144 (1 – cos2θ)
⇒ cos2θ – 130 cosθ + 169
= 144 – 144 cos2θ
⇒ 144 cos2θ + 25 cos2θ – 130 cosθ + 169 – 144 = 0
⇒ 169 cos2θ – 130 cosθ + 25 = 0
⇒ (13 cosθ – 5)2 = 0
⇒ 13 cosθ – 5 = 0
⇒ 13 cosθ = 5⇒ cosθ = 5 13
OR
5 cosθ + 12 sinθ = 13⇒ 5 cosθ + 12 sinθ = 1 13 13
∵ cos2θ + sin2θ = 1∴ cosθ = 5 13 Correct Option: B
5 cosθ + 12 sinθ = 13
⇒ 5 cosθ – 13 = 12 sinθ
On squaring both sides,
25 cos2θ + 169 – 130 cosθ
= 144 (1 – cos2θ)
⇒ cos2θ – 130 cosθ + 169
= 144 – 144 cos2θ
⇒ 144 cos2θ + 25 cos2θ – 130 cosθ + 169 – 144 = 0
⇒ 169 cos2θ – 130 cosθ + 25 = 0
⇒ (13 cosθ – 5)2 = 0
⇒ 13 cosθ – 5 = 0
⇒ 13 cosθ = 5⇒ cosθ = 5 13
OR
5 cosθ + 12 sinθ = 13⇒ 5 cosθ + 12 sinθ = 1 13 13
∵ cos2θ + sin2θ = 1∴ cosθ = 5 13
