Question
Evaluate the following: $\frac{\tan 42^{\circ}}{\cot 48^{\circ}}+\frac{\cos 33^{\circ}}{\sin 57^{\circ}}$
✓
Answer
$ \frac{\tan 42^{\circ}}{\cot 48^{\circ}}+\frac{\cos 33^{\circ}}{\sin 57^{\circ}}$
$ =\frac{\tan \left(90^{\circ}-48^{\circ}\right)}{\cot 48^{\circ}}-\frac{\cos \left(90^{\circ}-57^{\circ}\right)}{\sin 57^{\circ}}$
$ =\frac{\cot 48^{\circ}}{\cot 48^{\circ}}-\frac{\sin 57^{\circ}}{\sin 57^{\circ}}$
$ =1-1$
$ =0 .$
$ =\frac{\tan \left(90^{\circ}-48^{\circ}\right)}{\cot 48^{\circ}}-\frac{\cos \left(90^{\circ}-57^{\circ}\right)}{\sin 57^{\circ}}$
$ =\frac{\cot 48^{\circ}}{\cot 48^{\circ}}-\frac{\sin 57^{\circ}}{\sin 57^{\circ}}$
$ =1-1$
$ =0 .$
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