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If ${\left( {\frac{{\sin \theta }}{{\sin \phi }}} \right)^2} = \frac{{\tan \theta }}{{\tan \phi }} = 3,$ then the value of $\theta $ and $\phi $ are
$\theta = n\pi \pm \frac{\pi }{3},\,\phi = n\pi \pm \frac{\pi }{6}$
$\theta = n\pi - \frac{\pi }{3},\,\phi = n\pi - \frac{\pi }{6}$
$\theta = n\pi \pm \frac{\pi }{2},\,\phi = n\pi + \frac{\pi }{3}$
None of these
Solution
(a) ${\left( {\frac{{\sin \theta }}{{\sin \phi }}} \right)^2} = \frac{{\tan \theta }}{{\tan \phi }}$
$ \Rightarrow $ $\sin \theta .\cos \theta = \sin \phi \cos \phi $
$ \Rightarrow $ $\sin 2\theta = \sin 2\phi $
$2\theta = \pi – 2\phi $ $ \Rightarrow $ $\theta = \frac{\pi }{2} – \phi $
But $\frac{{\tan \theta }}{{\tan \phi }} = 3$ $ \Rightarrow $ $\frac{{\tan \theta }}{{\cot \theta }} = 3$ $ \Rightarrow $ ${\tan ^2}\theta = 3$
$ \Rightarrow $ $\theta = n\pi \pm \frac{\pi }{3}$, so that $\phi = n\pi \pm \frac{\pi }{6}$.
Trick : Check with the options for $n = 0,\,1$.
