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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If the expression ( 1 + tan x + ^ 2 x ) ( 1 - cot x + ^ 2 x ) is...

If the expression (1+tanx+tan⁡2x)(1−cotx+cot⁡2x)( 1 + tan x + \tan^{2} x ) ( 1 - cot x + \cot^{2} x ) is positive, then the complete set of values of xx is

Options

  1. A[0,π2]\left[0, \dfrac{\pi}{2}\right]
  2. B[0,π]\left[0, \pi\right]
  3. CR−{x=nπ2,n∈I}\mathbb{R} - \left\{x = \dfrac{n\pi}{2}, n \in I\right\}
  4. D[0,∞]\left[0, \infty\right]

Correct answer

C. R−{x=nπ2,n∈I}\mathbb{R} - \left\{x = \dfrac{n\pi}{2}, n \in I\right\}

Step-by-step solution

(1+tanx+tan⁡2x)(1+tan2x−tanx)tan⁡2x>0\dfrac{(1 + tan x + \tan^{2} x ) ( 1 + tan^{2}x - tan x )}{\tan^{2} x} > 0

⇒(1+tan⁡2x)2−tan⁡2xtan⁡2x>0⇒ \dfrac{\left(1 + \tan^{2} x\right)2 - \tan^{2} x}{\tan^{2} x} > 0

Since, 1+tan⁡2x>tan⁡2x,∀x∈R−{x=nπ2,n∈I}1 + \tan^{2} x > \tan^{2} x, \forall x\in \mathbb{R} - \left\{x = \dfrac{n\pi}{2}, n \in I\right\}

Hence, given expression is positive for all values of x∈R−{x=nπ2,n∈I}x\in \mathbb{R} - \left\{x = \dfrac{n\pi}{2}, n \in I\right\}
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