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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

If sin x + cos x = 7 2 , where x ∈ [0, (π)/(4)], then the value of...

If sinx+cosx=72sin x + cos x = \dfrac{\sqrt{7}}{2}, where x∈x \in\left[0, \dfrac{π}{4}\right],thenthevalueofthen the value oftan\dfrac{x}{2} is equal to

Options

  1. A3−73\dfrac{3 - \sqrt{7}}{3}
  2. B7−23\dfrac{\sqrt{7} - 2}{3}
  3. C4−74\dfrac{4 - \sqrt{7}}{4}
  4. D5−32\dfrac{5 - \sqrt{3}}{2}

Correct answer

B. 7−23\dfrac{\sqrt{7} - 2}{3}

Step-by-step solution

sinx+cosx=72sin x + cos x = \dfrac{\sqrt{7}}{2}
⇒2tanx21+tan2x2+1−tan2x21+tan2x2=72⇒ \dfrac{2tan\dfrac{x}{2}}{1 + tan^{2}\dfrac{x}{2}} + \dfrac{1 - tan^{2}\dfrac{x}{2}}{1 + tan^{2}\dfrac{x}{2}} = \dfrac{7}{2}
⇒4tanx2+2−2tan2x2=7+7tan2x2⇒ 4tan\dfrac{x}{2} + 2 - 2tan^{2}\dfrac{x}{2} = \sqrt{7} + \sqrt{7}tan^{2}\dfrac{x}{2}
⇒(2+7)tan2x2−4tanx2+7−2=0⇒ \left(2 + \sqrt{7}\right)tan^{2}\dfrac{x}{2} - 4tan\dfrac{x}{2} + \sqrt{7} - 2 = 0
⇒tanx2=4±16−4(2+7)(7−2)2(2+7)⇒ tan\dfrac{x}{2} = \dfrac{4 \pm \sqrt{16 - 4\left(2 + \sqrt{7}\right)\left(\sqrt{7} - 2\right)}}{2\left(2 + \sqrt{7}\right)}
=4±16−4×32(2+7)= \dfrac{4 \pm \sqrt{16 - 4 \times 3}}{2\left(2 + \sqrt{7}\right)}
tanx2=2±12+7=12+7or32+7tan\dfrac{x}{2} = \dfrac{2 \pm 1}{2 + \sqrt{7}} = \dfrac{1}{2 + \sqrt{7}}or\dfrac{3}{2 + \sqrt{7}}
tanx2=7−23tan\dfrac{x}{2} = \dfrac{\sqrt{7} - 2}{3} or 7−\sqrt{7} -2<br/>But,asx2∈[0,π8]<br/><br />But, as \dfrac{x}{2} \in \left[0, \dfrac{\pi}{8}\right]<br />⇒ tanx2=7−23tan\dfrac{x}{2} = \dfrac{\sqrt{7} - 2}{3}
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