Question
Evaluate the following$: \sin 28^\circ \sec 62^\circ + \tan 49^\circ \tan 41^\circ $
✓
Answer
$ \sin 28^{\circ} \sec 62^{\circ}+\tan 49^{\circ} \tan 41^{\circ}$
$ =\sin 28^{\circ} \sec \left(90^{\circ}-28^{\circ}\right)+\tan 49^{\circ} \tan \left(90^{\circ}-49^{\circ}\right)$
$ =\sin 28^{\circ} \operatorname{cosec} 28^{\circ}+\tan 49^{\circ} \cot 49^{\circ}$
$ =\sin 28^{\circ} \times \frac{1}{\sin 28^{\circ}}+\tan 49^{\circ} \times \frac{1}{\tan 49^{\circ}}$
$ =1+1$
$ =2 .$
$ =\sin 28^{\circ} \sec \left(90^{\circ}-28^{\circ}\right)+\tan 49^{\circ} \tan \left(90^{\circ}-49^{\circ}\right)$
$ =\sin 28^{\circ} \operatorname{cosec} 28^{\circ}+\tan 49^{\circ} \cot 49^{\circ}$
$ =\sin 28^{\circ} \times \frac{1}{\sin 28^{\circ}}+\tan 49^{\circ} \times \frac{1}{\tan 49^{\circ}}$
$ =1+1$
$ =2 .$
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