Question
Without using trigonometric identity, show that :
$
\sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ}=2
$
$
\sin 42^{\circ} \sec 48^{\circ}+\cos 42^{\circ} \cos e c 48^{\circ}=2
$
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(i) (6, -3)
(ii) (-1, 0)
(iii) (-8, -2)