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Prove that:$ \sin 48^{\circ} \sec 42^{\circ}+\cos 48^{\circ} \operatorname{cosec} 42^{\circ}=2 $
To do:
We have to prove that $\sin 48^{\circ} \sec 42^{\circ}+\cos 48^{\circ} \operatorname{cosec} 42^{\circ}=2$.
Solution:
We know that,
$sin\ (90^{\circ}- \theta) = cos\ \theta$
$cos\ (90^{\circ}- \theta) = sin\ \theta$
$cos\ \theta \times \sec\ \theta=1$
$sin\ \theta \times \operatorname{cosec}\ \theta=1$
Therefore,
$\sin 48^{\circ} \sec 42^{\circ}+\cos 48^{\circ} \operatorname{cosec} 42^{\circ}=\sin (90^{\circ}- 42^{\circ})\sec 42^{\circ} + \cos (90^{\circ}- 42^{\circ})\operatorname{cosec} 42^{\circ}$
$=\cos 42^{\circ} \sec 42^{\circ} + \sin 42^{\circ} \operatorname{cosec} 42^{\circ}$
$=1+1$
$=2$
Hence proved.
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