Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
(N/A) $(i)$ We know that $\tan(90^{\circ} - \theta) = \cot \theta$.
$\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = \tan(90^{\circ} - 42^{\circ}) \tan(90^{\circ} - 67^{\circ}) \tan 42^{\circ} \tan 67^{\circ}$
$= \cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$= (\cot 42^{\circ} \tan 42^{\circ}) (\cot 67^{\circ} \tan 67^{\circ})$
$= (1)(1) = 1$.
$(ii)$ We know that $\cos(90^{\circ} - \theta) = \sin \theta$ and $\sin(90^{\circ} - \theta) = \cos \theta$.
$\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = \cos(90^{\circ} - 52^{\circ}) \cos(90^{\circ} - 38^{\circ}) - \sin 38^{\circ} \sin 52^{\circ}$
$= \sin 52^{\circ} \sin 38^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$.
$\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = \tan(90^{\circ} - 42^{\circ}) \tan(90^{\circ} - 67^{\circ}) \tan 42^{\circ} \tan 67^{\circ}$
$= \cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$= (\cot 42^{\circ} \tan 42^{\circ}) (\cot 67^{\circ} \tan 67^{\circ})$
$= (1)(1) = 1$.
$(ii)$ We know that $\cos(90^{\circ} - \theta) = \sin \theta$ and $\sin(90^{\circ} - \theta) = \cos \theta$.
$\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = \cos(90^{\circ} - 52^{\circ}) \cos(90^{\circ} - 38^{\circ}) - \sin 38^{\circ} \sin 52^{\circ}$
$= \sin 52^{\circ} \sin 38^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$.
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