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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

NTA Abhyas 2020 · Mathematics · Trigonometric Ratios & Identities

If 4sin 27 ^ o = α - β , then the value of α + β is

If 4sin27o=α−β4sin {27}^{o} = \sqrt{\alpha} - \sqrt{\beta}, then the value of α+β\alpha + \beta is

Options

  1. A55
  2. B33
  3. C88
  4. D22

Correct answer

C. 88

Step-by-step solution

(cos27o+sin27o)2=1+2sin27ocos27o=1+sin54o=1+cos36o{\left(cos {27}^{o} + sin {27}^{o}\right)}^{2} = 1 + 2sin {27}^{o}cos {27}^{o} = 1 + sin {54}^{o} = 1 + cos {36}^{o}
⇒cos27o+sin27o=1+cos36o⇒ cos {27}^{o} + sin {27}^{o} = \sqrt{1 + cos {36}^{o}}
Similarly, cos27o−sin27o=1−cos36ocos {27}^{o} - sin {27}^{o} = \sqrt{1 - cos {36}^{o}} (since, cos27o>sin27ocos {27}^{o} > sin {27}^{o} )
On subtracting, we get,
2sin27o=1+cos36o−1−cos36o2sin {27}^{o} = \sqrt{1 + cos {36}^{o}} - \sqrt{1 - cos {36}^{o}}
=1+(5+14)−1−(5+14)= \sqrt{1 + \left(\dfrac{\sqrt{5} + 1}{4}\right)} - \sqrt{1 - \left(\dfrac{\sqrt{5} + 1}{4}\right)}
=5+52−3−52= \dfrac{\sqrt{5 + \sqrt{5}}}{2} - \dfrac{\sqrt{3 - \sqrt{5}}}{2}
⇒4sin27o=5+5−3−5⇒ 4sin {27}^{o} = \sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}
⇒α=5+5⇒ \alpha = 5 + \sqrt{5} and β=3−5\beta = 3 - \sqrt{5}
⇒α+β=8⇒ \alpha + \beta = 8
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