Question
Prove the following$: sin58^\circ sec32^\circ + cos58^\circ \operatorname{cosec}32^\circ = 2$
✓
Answer
$ \text { L.H.S. }$
$ =\sin 58^{\circ} \sec 32^{\circ}+\cos 58^{\circ} \operatorname{cosec} 32^{\circ}$
$ =\sin \left(90^{\circ}-32^{\circ}\right) \times \frac{1}{\cos 32^{\circ}}+\cos \left(90^{\circ}-32^{\circ}\right) \times \frac{1}{\sin 32^{\circ}}$
$ =\cos 32^{\circ} \times \frac{1}{\cos 32^{\circ}}+\sin 32^{\circ} \times \frac{1}{\sin 32^{\circ}}$
$ =1+1$
$ =2$
$ =\text { R.H.S. }$
$ =\sin 58^{\circ} \sec 32^{\circ}+\cos 58^{\circ} \operatorname{cosec} 32^{\circ}$
$ =\sin \left(90^{\circ}-32^{\circ}\right) \times \frac{1}{\cos 32^{\circ}}+\cos \left(90^{\circ}-32^{\circ}\right) \times \frac{1}{\sin 32^{\circ}}$
$ =\cos 32^{\circ} \times \frac{1}{\cos 32^{\circ}}+\sin 32^{\circ} \times \frac{1}{\sin 32^{\circ}}$
$ =1+1$
$ =2$
$ =\text { R.H.S. }$
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