MCQ
$\sin 12^\circ \sin 24^\circ \sin 48^\circ \sin 84^\circ = $
- ✓$\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ $
- B$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ $
- C$\frac{3}{{15}}$
- DNone of these
$ = \frac{1}{4}\,(2\,\,\sin \,\,{12^o}\,\sin \,\,{48^o})\,\,(2\,\,\sin \,\,{24^o}\,\,\sin \,\,{84^o})$
$ = \frac{1}{2}(\cos \,\,{36^o} - \cos \,\,{60^o})\,\,(\cos \,\,{60^o} - \cos \,\,{108^o})$
$ = \frac{1}{4}\,\left( {\cos \,\,{{36}^o} - \frac{1}{2}} \right)\,\,\left( {\frac{1}{2} + \sin \,\,{{18}^o}} \right)$
$ = \frac{1}{4}\left\{ {\frac{1}{4}(\sqrt 5 + 1) - \frac{1}{2}} \right\}\,\left\{ {\frac{1}{2} + \frac{1}{4}(\sqrt 5 - 1)} \right\} = \frac{1}{{16}}$
and $\cos \,\,{20^o}\,\cos \,\,{40^o}\,\,\cos \,\,60\,\,\cos \,\,{80^o}$
$ = \frac{1}{2}[\cos \,({60^o} - {20^o})\,\cos \,\,{20^o}\,\cos \,({60^o} + {20^o})]$
$ = \frac{1}{2}\,\left[ {\frac{1}{4}\cos \,\,3\,\,({{20}^o})} \right] = \frac{1}{8}\cos \,\,{60^o} = \frac{1}{2} \times \frac{1}{8} = \frac{1}{{16}}$.
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$(A)$ the equation of the hyperbola is $\frac{x^2}{3}-\frac{y^2}{2}=1$
$(B)$ a focus of the hyperbola is $(2,0)$
$(C)$ the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$
$(D)$ the equation of the hyperbola is $x^2-3 y^2=3$