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Trigonometrical Equations
easy
The most general value of $\theta $ which will satisfy both the equations $\sin \theta = - \frac{1}{2}$ and $\tan \theta = \frac{1}{{\sqrt 3 }}$ is
A
$n\pi + {( - 1)^n}\frac{\pi }{6}$
B
$n\pi + \frac{\pi }{6}$
C
$2n\pi \pm \frac{\pi }{6}$
D
None of these
Solution
(d) $\sin \theta = – \frac{1}{2} = \sin \left( { – \frac{\pi }{6}} \right) = \sin {\rm{ }}\left( {\pi + \frac{\pi }{6}} \right)$
$\tan \theta = \frac{1}{{\sqrt 3 }} = \tan \left( {\frac{\pi }{6}} \right) = \tan \left( {\pi + \frac{\pi }{6}} \right)$
$\Rightarrow \theta = \left( {\pi + \frac{\pi }{6}} \right)$
Hence general value of $\theta $ is $2n\pi + \frac{{7\pi }}{6}$.
Standard 11
Mathematics
