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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If (k + 1)θ = tan θ , then the set of all the values of θ is&

If tan⁡(k+1)θ=tanθ\tan \left(k + 1\right)\theta = tan \theta, then the set of all the values of θ\theta is&

Options

  1. Anπ:n∈I{ n\pi: n \in I }
  2. B\left\{\dfrac{\text{n} π}{2} \text{:} \text{n} ∈ \text{I}\right\}
  3. C{nπk:n∈I}\left\{\dfrac{\text{n} \pi}{\text{k}} \text{:} \text{n} \in \text{I}\right\}
  4. D\left\{\dfrac{\text{n} π}{4} \text{:} \text{n} ∈ \text{I}\right\}

Correct answer

C. {nπk:n∈I}\left\{\dfrac{\text{n} \pi}{\text{k}} \text{:} \text{n} \in \text{I}\right\}

Step-by-step solution

As we know the general solution of the equation

tan⁡x=tanαisgivenbyx=nπ+α,n∈I\tan x = tan\alpha\text{is}\text{given}\text{by}x = n\pi + \alpha, n \in I

Hence the solution of the equation, tan(k+1)θ=tanθ\text{tan} \left(\text{k} + 1\right) \theta = \text{tan} \theta

is given by \left(\text{k} + 1\right) \theta = \text{n} \pi + \theta ⇒ \text{k} \theta = \text{n} \pi ,\text{n} \in I<br/><br/><br /><br />∴ \theta \in \dfrac{n\pi}{k}: n \in I&
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