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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

Three positive acute angles α, β and γ satisfy the relation tan...

Three positive acute angles α,β\alpha, \beta and γ\gamma satisfy the relation tanβ2=13cotα2tan \dfrac{\beta}{2} = \dfrac{1}{3}cot \dfrac{\alpha}{2} and cot \dfrac{\gamma}{2} = \dfrac{1}{2} \left(3tan \dfrac{\alpha}{2} + cot \dfrac{\alpha}{2}\right). Then, the value of\alpha + \beta + \gamma isequaltoequal to

Options

  1. Aπ\pi
  2. B2π2\pi
  3. Cπ2\dfrac{\pi}{2}
  4. D3π2\dfrac{3\pi}{2}

Correct answer

A. π\pi

Step-by-step solution

We have,& tanα2tanβ2=13tan \dfrac{\alpha}{2}tan \dfrac{\beta}{2} = \dfrac{1}{3}
⇒1−tanα2tanβ2=23⇒ 1 - tan \dfrac{\alpha}{2}tan \dfrac{\beta}{2} = \dfrac{2}{3}
Now, tan⁡(α+β2)=tanα2+tanβ21−tanα2tanβ2=tanα2+tanβ223\tan \left(\dfrac{\alpha + \beta}{2}\right) = \dfrac{tan \dfrac{\alpha}{2} + tan \dfrac{\beta}{2}}{1 - tan \dfrac{\alpha}{2}tan \dfrac{\beta}{2}} = \dfrac{tan \dfrac{\alpha}{2} + tan \dfrac{\beta}{2}}{\dfrac{2}{3}}
⇒ \tan \left(\dfrac{\alpha + \beta}{2}\right) = \dfrac{3}{2} \left( tan α2+\dfrac{\alpha}{2} + tan \dfrac{\beta}{2}\right)<br />=\dfrac{3}{2} tan α2+12(3tanβ2)=32\dfrac{\alpha}{2} + \dfrac{1}{2}\left(3tan \dfrac{\beta}{2}\right) = \dfrac{3}{2} tan α2+12\dfrac{\alpha}{2} + \dfrac{1}{2} cot α2<br/>\dfrac{\alpha}{2}<br />= cot γ2<br/>\dfrac{\gamma}{2}<br />⇒ \tan \left(\dfrac{\alpha + \beta}{2}\right) =cotcot\dfrac{\gamma}{2} = \dfrac{1}{tan \dfrac{\gamma}{2}}
⇒1−tan⁡(α+β2)⇒ 1 - \tan \left(\dfrac{\alpha + \beta}{2}\right)⋅ tan γ2=0\dfrac{\gamma}{2} =0
⇒(α+β2)+γ2=π2⇒ \left(\dfrac{\alpha + \beta}{2}\right) + \dfrac{\gamma}{2} = \dfrac{\pi}{2}⇒ \alpha + \beta + \gamma = \pi$$
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