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, $\frac{\sin 50^{\circ}}{\cos 40^{\circ}}+\frac{\operatorname{cosec} 40^{\circ}}{\sec 50^{\circ}}-4 \cos 50^{\circ} \operatorname{cosec} 40^{\circ}=-2$.   

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Simply Easy Learning

Updated on: 10-Oct-2022

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