Question
Show that $: \sin 42^\circ \sec 48^\circ + \cos 42^\circ \operatorname{cosec} 48^\circ = 2.$
✓
Answer
$\text{L.H.S.}$
$= \sin 42^\circ \sec 48^\circ + \cos 42^\circ \operatorname{cosec} 48^\circ$
$=\sin \left(90^{\circ}-48^{\circ}\right) \times \frac{1}{\cos 48^{\circ}}+\cos \left(90^{\circ}-48^{\circ}\right) \times \frac{1}{\sin 48^{\circ}}$
$=\cos 48^{\circ} \times \frac{1}{\cos 48^{\circ}}+\sin 48^{\circ} \times \frac{1}{\sin 48^{\circ}}$
$= 1 + 1$
$= 2$
$=\text{ R.H.S.}$
$= \sin 42^\circ \sec 48^\circ + \cos 42^\circ \operatorname{cosec} 48^\circ$
$=\sin \left(90^{\circ}-48^{\circ}\right) \times \frac{1}{\cos 48^{\circ}}+\cos \left(90^{\circ}-48^{\circ}\right) \times \frac{1}{\sin 48^{\circ}}$
$=\cos 48^{\circ} \times \frac{1}{\cos 48^{\circ}}+\sin 48^{\circ} \times \frac{1}{\sin 48^{\circ}}$
$= 1 + 1$
$= 2$
$=\text{ R.H.S.}$
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