MCQ
$\frac{\cos 53^{\circ}}{\sin 37^{\circ}}=$ ?
- A$0$
- B$\frac{1}{2}$
- ✓$1$
- DNone of these
✓
Answer
Correct option: C.
$1$
$\frac{\cos 53^{\circ}}{\sin 37^{\circ}}=\frac{\cos \left(90^{\circ}-37^{\circ}\right)}{\sin 37^{\circ}}$
$=\frac{\sin 37^{\circ}}{\sin 37^{\circ}}$
$=1 ([\because \cos (90-\theta)=\sin \theta])$
$=\frac{\sin 37^{\circ}}{\sin 37^{\circ}}$
$=1 ([\because \cos (90-\theta)=\sin \theta])$
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