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Evaluate:
$ \frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ} $
Given:
\( \frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ} \)
To do:
We have to evaluate \( \frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ} \).
Solution:
We know that,
$sin\ (90^{\circ}- \theta) = cos\ \theta$
$\operatorname{cosec}\ (90^{\circ}- \theta) =\sec\ \theta$
$cos\ (90^{\circ}- \theta) = sin\ \theta$
$sin\ \theta \times \operatorname{cosec}\ \theta=1$
$\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}=\frac{\sin (90^{\circ}-40^{\circ})}{\cos 40^{\circ}}+\frac{\operatorname{cosec} (90^{\circ}-50^{\circ})}{\sec 50^{\circ}}-4 \cos (90^{\circ}-40^{\circ}) \operatorname{cosec} 40^{\circ}$
$=\frac{\cos 40^{\circ}}{\cos 40^{\circ}}+\frac{\sec 50^{\circ}}{\sec 50^{\circ}}-4 \sin 40^{\circ} \operatorname{cosec} 40^{\circ}$
$=1+1-4(1)$
$=2-4$
$=-2$
Hence, $\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}=-2$.