$\frac{\cos 12^{\circ}-\sin 12^{\circ}}{\cos 12^{\circ}+\sin 12^{\circ}}+\frac{\sin 147^{\circ}}{\cos 147^{\circ}}=$

Question

Aicharya · Accepted Answer

$\frac{\cos 12^{\circ}-\sin 12^{\circ}}{\cos 12^{\circ}+\sin 12^{\circ}}+\frac{\sin 147^{\circ}}{\cos 147^{\circ}}$ $=\frac{\cos 12^{\circ}-\cos 78^{\circ}}{\cos 12^{\circ}+\cos 78^{\circ}}+	an 147^{\circ}$ $=\frac{-2 \sin 45^{\circ} \sin \left(-33^{\circ}ight)}{2 \cos 45^{\circ} \cos 33^{\circ}}+	an \left(180^{\circ}-33^{\circ}ight)=	an 45^{\circ} 	an 33^{\circ}-	an 33^{\circ}$ $=	an 33^{\circ}-	an 33^{\circ}=0$

$\frac{\cos 12^{\circ}-\sin 12^{\circ}}{\cos 12^{\circ}+\sin 12^{\circ}}+\frac{\sin 147^{\circ}}{\cos…

Solution