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Prove that:$ \frac{\sin 70^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\sec 70^{\circ}}-2 \cos 70^{\circ} \operatorname{cosec} 20^{\circ}=0 $
To do:
We have to prove that $\frac{\sin 70^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\sec 70^{\circ}}-2 \cos 70^{\circ} \operatorname{cosec} 20^{\circ}=0$.
Solution:
We know that,
$sin\ (90^{\circ}- \theta) = cos\ \theta$
$sec\ (90^{\circ}- \theta) = \operatorname{cosec}\ \theta$
$cos\ (90^{\circ}- \theta) = sin\ \theta$
$sin\ \theta \times \operatorname{cosec}\ \theta=1$
Therefore,
$\frac{\sin 70^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\sec 70^{\circ}}-2 \cos 70^{\circ} \operatorname{cosec} 20^{\circ}=\frac{\sin (90^{\circ}- 20^{\circ})}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\sec (90^{\circ}- 20^{\circ})}-2 \cos (90^{\circ}- 20^{\circ}) \operatorname{cosec} 20^{\circ}$
$=\frac{\cos 20^{\circ}}{\cos 20^{\circ}}+\frac{\operatorname{cosec} 20^{\circ}}{\operatorname{cosec} 20^{\circ}}-2 \sin 20^{\circ} \operatorname{cosec} 20^{\circ}$
$=1+1-2(1)$
$=2-2$
$=0$
Hence proved.
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