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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 cos α + β = a , α + β = b and α - β = 2 θ , then ( 3 θ)/( θ) is...

\text{If cos} \alpha + \text\cos \beta = \text{a} \text{,} \text\sin \alpha + \text\sin \beta = \text{b} \text{and} \alpha - \beta = 2 \theta \text{,} \text{then} \dfrac{\text\cos 3 \theta}{\text\cos \theta} is equal to

Options

  1. Aa2+b2−2a^{2} + b^{2} - 2
  2. Ba2+b2−3a^{2} + b^{2} - 3
  3. C3−a2−b23 - a^{2} - b^{2}
  4. Da2+b24\dfrac{a^{2} + b^{2}}{4}

Correct answer

B. a2+b2−3a^{2} + b^{2} - 3

Step-by-step solution

\dfrac{\text{cos} 3 \theta}{\text{cos} \theta} = 4 \cos^{2~} \theta - 3 = 2 \left(1 + \text{cos} 2 \theta\right) - 3<br/><br/><br /><br />= 2 \text{cos} 2 \theta - 1 = 2 \text{cos} \left(\alpha - \beta\right) - 1<br /><br />Now,~\left({& cos}^{2& } \alpha + \sin^{2& } \alpha\right) + \left(\cos^{2& } \beta + \sin^{2& } \beta\right) + 2 \text{cos} \left(\alpha - \beta\right) = a^{2} + b^{2}

⇒2cos(α−β)=a2+b2−2⇒ 2 \text{cos} \left(\alpha - \beta\right) = a^{2} + b^{2} - 2

⇒cos3θcosθ=a2+b2−3⇒ \dfrac{\text{cos} 3 \theta}{\text{cos} \theta} = a^{2} + b^{2} - 3
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