Question.
Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
Show that
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0.
Solution:
(i) $\mathrm{LHS}=\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$=\tan 48^{\circ} \times \tan 23^{\circ} \times \tan \left(90^{\circ}-48^{\circ}\right)$ $\times \tan \left(90^{\circ}-23^{\circ}\right)$
$=\tan 48^{\circ} \times \tan 23^{\circ} \times \cot 48^{\circ} \times \cot 23^{\circ}$
$=\tan 48^{\circ} \times \tan 23^{\circ} \times \frac{1}{\tan 48^{\circ}} \times \frac{1}{\tan 23^{\circ}}=1$
$\therefore \quad \mathrm{LHS}=\mathrm{RHS} .$
(ii) LHS = $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}$
$=\cos \left(90^{\circ}-52^{\circ}\right) \cos 52^{\circ}-\sin \left(90^{\circ}-52^{\circ}\right) \sin 52^{\circ}$
$=\sin 52^{\circ} \cos 52^{\circ}-\cos 52^{\circ} \sin 52^{\circ}=0$
(i) $\mathrm{LHS}=\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$=\tan 48^{\circ} \times \tan 23^{\circ} \times \tan \left(90^{\circ}-48^{\circ}\right)$ $\times \tan \left(90^{\circ}-23^{\circ}\right)$
$=\tan 48^{\circ} \times \tan 23^{\circ} \times \cot 48^{\circ} \times \cot 23^{\circ}$
$=\tan 48^{\circ} \times \tan 23^{\circ} \times \frac{1}{\tan 48^{\circ}} \times \frac{1}{\tan 23^{\circ}}=1$
$\therefore \quad \mathrm{LHS}=\mathrm{RHS} .$
(ii) LHS = $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}$
$=\cos \left(90^{\circ}-52^{\circ}\right) \cos 52^{\circ}-\sin \left(90^{\circ}-52^{\circ}\right) \sin 52^{\circ}$
$=\sin 52^{\circ} \cos 52^{\circ}-\cos 52^{\circ} \sin 52^{\circ}=0$
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