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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If x - x = - (5)/(4) , where (π)/(2) < x < (3π)/(4) , then...

If cos⁡x−sin⁡x=−54\cos x - \sin x = - \dfrac{5}{4}, where π2<x<3π4\dfrac{π}{2} < x < \dfrac{3\pi}{4}, then cotx2cot\dfrac{x}{2}& is equal to

Options

  1. A4−79\dfrac{4 - \sqrt{7}}{9}
  2. B88
  3. C−8- 8
  4. D4+79\dfrac{4 + \sqrt{7}}{9}

Correct answer

D. 4+79\dfrac{4 + \sqrt{7}}{9}

Step-by-step solution

Given, 4(cosx−sinx)=−54\left(cos x - sin x\right) = - 5
4(1−tan2x21+tan2x2−2tanx21+tan2x2)=−54\left(\dfrac{1 - tan^{2}\dfrac{x}{2}}{1 + tan^{2}\dfrac{x}{2}} - \dfrac{2tan \dfrac{x}{2}}{1 + tan^{2}\dfrac{x}{2}}\right) =- 5
⇒4−4tan⁡2x2−8tanx2=−5−5tan⁡2x2⇒ 4 - 4\tan^{2} \dfrac{x}{2} - 8tan \dfrac{x}{2} = - 5 - 5\tan^{2} \dfrac{x}{2}
⇒tan2x2−8tanx2+9=0⇒ tan^{2}\dfrac{x}{2} - 8tan \dfrac{x}{2} + 9 = 0
⇒9cot⁡2x2−8cotx2+1=0⇒ 9\cot^{2} \dfrac{x}{2} - 8cot \dfrac{x}{2} + 1 = 0
⇒cotx2=4±79⇒ cot\dfrac{x}{2} = \dfrac{4 \pm \sqrt{7}}{9}
Since, π4<x2<3π8⇒2−1<cotx2<1\dfrac{\pi}{4} < \dfrac{x}{2} < \dfrac{3\pi}{8} ⇒ \sqrt{2} - 1 < cot\dfrac{x}{2} < 1
⇒cotx2=4+79⇒ cot\dfrac{x}{2} = \dfrac{4 + \sqrt{7}}{9}
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