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Trigonometrical Equations
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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
$\frac{\pi}{15}$
B
$\frac{5\pi}{18}$
C
$\frac{5\pi}{12}$
D
$\frac{17\pi}{18}$
Solution
$\frac{\tan (\theta+\alpha)}{\tan (\theta-\alpha)}=\frac{3}{1}$
by using componendo and dividendo we get
$\sin 2 \theta=2 \sin 2 \alpha$
this equation has no solution if $|\sin 2 \alpha|>\frac{1}{2}$
Standard 11
Mathematics
