UP Board Important Question
Mathematics (Trigonometric Equations)
Class – XII
[Set-3]
निम्निलिखित समीकरण को हल करके θ के व्यापक मान (Comprehensive Value) ज्ञात कीजिये |
Q1. √3.tanθ – 1 =0
Solution:
√3.tanθ – 1 =0
=> √3.tanθ = 1
=> tanθ = 1/√3
=> tanθ = tanπ/6
So θ of Comprehensive Value = nπ + π/6
Q2. √2 . sinθ – 1 = 0
Q3. cosθ = √3/2
Solution:
cosθ = √3/2
=> cosθ = cosπ/6
So θ of Comprehensive Value = 2nπ ± π/3
Q4. cos2θ = 1/4
Q5. cos2θ = cos2θ
Solution:
=> cos2θ = cos2θ
=> 2 cos2θ - 1 = cos2θ
=> cos2θ - 1 = 0
=> cos2θ = 12
=> cos2θ = cos20
So θ of Comprehensive Value = nπ ± 0 = nπ
Q6. cos2θ = 1/2
Q7. 2 (cos2θ - sin2θ) = 1
Solution:
=> 2 (cos2θ - sin2θ) = 1
=> cos2θ - sin2θ = ½
=> cos2θ = 1/2
=> cos2θ = cosπ/3
=> 2θ = 2nπ ± π/3
So θ of Comprehensive Value = nπ ± π/6
Q8. tanθ = cotθ
Solution:
=> tanθ = cotθ
=> tanθ = 1/ tanθ
=> tan2θ = 1
=> tan2θ = 12
=> tan2θ = tan2 π/4
So θ of Comprehensive Value = nπ ± π/4
Q9. cos2θ – sinθ . cosθ – ½ = 0
Solution:
=> cos2θ – sinθ . cosθ – ½ = 0
=> 2cos2θ – 2sinθ . cosθ - 1 = 0
=> (2cos2θ – 1) – 2sinθ . cosθ = 0
=> cos2θ – sin2θ = 0
=> cos2θ = sin2θ
=> sin2θ/ cos2θ = 1
=> tan2θ = tanπ/4
=> 2θ = nπ ± π/4
So θ of Comprehensive Value = nπ/2 ± π/8
Q10. 1 + cotθ = cosecθ
Q11. √2 . (sinθ + cosθ) = 1
Q12. tanθ + √2 . secθ = 1
Q13. sin2θ – 2 cosθ + ¼ = 0
Q14. 2 sin2θ + 3 cosθ = 0
Solution:
=> 2 sin2θ + 3 cosθ = 0
=> 2 (1 - cos2θ) + 3 cosθ = 0
=> 2 – 2cos2θ + 3 cosθ = 0
=> 2cos2θ - 3 cosθ - 2= 0
=> 2cos2θ - 4cosθ + cosθ - 2= 0
=> 2 cosθ (cosθ – 2) + (cosθ – 2)= 0
=> (cosθ – 2) (2cosθ + 1) = 0
=> cosθ – 2 = 0 OR 2cosθ + 1 = 0
If cosθ – 2 = 0 then cosθ = 2 > 1
But cosθ = 2 is invalid Because value of cosθ can’t grater then 1.
There for 2cosθ + 1 = 0 => cosθ = - 1/2 => cosθ = cos2π/3
So θ of Comprehensive Value = 2nπ ± 2π/3
Q15. 2 cos2θ = 1 + sinθ