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Trigonometrical Equations
easy
यदि $\sin \theta = \sqrt 3 \cos \theta , - \pi < \theta < 0$, तो $\theta = $
A
$ - \frac{{5\pi }}{6}$
B
$ - \frac{{4\pi }}{6}$
C
$\frac{{4\pi }}{6}$
D
$\frac{{5\pi }}{6}$
Solution
$\tan \theta = \sqrt 3 = \tan \frac{\pi }{3}$
$\Rightarrow \theta = n\pi + \frac{\pi }{3}$
$ – \pi < \theta < 0$ के लिये, $\,n = – 1$ रखने पर,
$\theta = – \pi + \frac{\pi }{3} = \frac{{ – 2\pi }}{3};k{\rm{ }}\frac{{ – 4\pi }}{6}$.
Standard 11
Mathematics
