Prove that:$ \frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}=2 $

To do:

We have to prove that $\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}=2$.

Solution:  

We know that,

$cos\ (90^{\circ}- \theta) = sin\ \theta$

$sin\ \theta \times \operatorname{cosec}\ \theta=1$

Therefore,

$\frac{\cos 80^{\circ}}{\sin 10^{\circ}}+\cos 59^{\circ} \operatorname{cosec} 31^{\circ}=\frac{\cos (90^{\circ}- 10^{\circ})}{\sin 10^{\circ}}+\cos (90^{\circ}- 31^{\circ}) \operatorname{cosec} 31^{\circ}$

$=\frac{\sin 10^{\circ}}{\sin 10^{\circ}}+\sin 31^{\circ} \operatorname{cosec} 31^{\circ}$

$=1+1$

$=2$

Hence proved.  

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Updated on: 10-Oct-2022

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