Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$
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) $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$=\tan \left(90^{\circ}-42^{\circ}\right) \tan \left(90^{\circ}-67^{\circ}\right) \tan 42^{\circ} \tan 67^{\circ}$
$=\cot 42^{*} \cot 67^{*} \tan 42^{*} \tan 67^{\circ}$
$=\left(\cot 42^{\circ} \tan 42^{\circ}\right)\left(\cot 67^{*} \tan 67^{\circ}\right)$
$=(1)(1)$
$=1$
(ii) $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}$
$=\cos \left(90^{\circ}-52^{\circ}\right) \cos \left(90^{\circ}-38^{\circ}\right)-\sin 38^{\circ} \sin 52^{\circ}$
$=\sin 52^{*} \sin 38^{\circ}-\sin 38^{\circ} \sin 52^{*}$
$=0$
Similar Questions
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