Question
Evaluate:
$\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\cos e c^2 10^{\circ}-\tan ^2 80^{\circ}}$
$\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\cos e c^2 10^{\circ}-\tan ^2 80^{\circ}}$
✓
Answer
$\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\cos e c^2 10^{\circ}-\tan ^2 80^{\circ}}$
$\frac{\sin 35^{\circ} \cdot \cos \left(90^{\circ}-35^{\circ}\right)+\cos 35^{\circ} \cdot \sin \left(90^{\circ}-35^{\circ}\right)}{\cos e c^2\left(90^{\circ}-80^{\circ}\right)-\tan ^2 80^{\circ}} $
$ =\frac{\sin 35^{\circ} \cdot \sin 35^{\circ}+\cos 35^{\circ} \cdot \cos 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}} $
$ =\frac{\sin ^2 35^{\circ}+\cos ^2 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}}=\frac{1}{1}=1$
$\frac{\sin 35^{\circ} \cdot \cos \left(90^{\circ}-35^{\circ}\right)+\cos 35^{\circ} \cdot \sin \left(90^{\circ}-35^{\circ}\right)}{\cos e c^2\left(90^{\circ}-80^{\circ}\right)-\tan ^2 80^{\circ}} $
$ =\frac{\sin 35^{\circ} \cdot \sin 35^{\circ}+\cos 35^{\circ} \cdot \cos 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}} $
$ =\frac{\sin ^2 35^{\circ}+\cos ^2 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}}=\frac{1}{1}=1$
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