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