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STD 11 - Trigonometrical equations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - Trigonometrical equations4 Marks
MCQ
If equation in variable $\theta, 3 tan(\theta -\alpha) = tan(\theta + \alpha)$, (where $\alpha$ is constant) has no real solution, then $\alpha$ can be (wherever $tan(\theta - \alpha)$ & $tan(\theta + \alpha)$ both are defined)
  • A
    $\frac{\pi}{15}$
  • ✓
    $\frac{5\pi}{18}$
  • C
    $\frac{5\pi}{12}$
  • D
    $\frac{17\pi}{18}$
✓

Answer

Correct option: B.
$\frac{5\pi}{18}$
b
$\frac{\tan (\theta+\alpha)}{\tan (\theta-\alpha)}=\frac{3}{1}$

by using componendo and dividendo we get

$\sin 2 \theta=2 \sin 2 \alpha$

this equation has no solution if $|\sin 2 \alpha|>\frac{1}{2}$

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