計算
\( \frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ} \)
已知
\( \frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ} \)
要求
我們需要計算\( \frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ} \).
解:
我們知道:
sin (90° - θ) = cos θ
cosec (90° - θ) = sec θ
cos (90° - θ) = sin θ
sin θ × cosec θ = 1
$\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}=\frac{\sin (90^{\circ}-40^{\circ})}{\cos 40^{\circ}}+\frac{\operatorname{cosec} (90^{\circ}-50^{\circ})}{\sec 50^{\circ}}-4 \cos (90^{\circ}-40^{\circ}) \operatorname{cosec} 40^{\circ}$
$=\frac{\cos 40^{\circ}}{\cos 40^{\circ}}+\frac{\sec 50^{\circ}}{\sec 50^{\circ}}-4 \sin 40^{\circ} \operatorname{cosec} 40^{\circ}$
$=1+1-4(1)$
$=2-4$
$=-2$
因此,$\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}=-2$