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FInd the value of sin 28/cos 62

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$\frac{\sin 28^{o}}{\cos 62^{o}}$

By using the identity of $\sin (90^o-\theta) = \cos \theta$

$=\frac{\sin (90^0-62^0)}{\cos 62^0}$
= $\frac{\cos 62^0}{\cos 62^0} = 1$

Posted by

Ravindra Pindel

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