1.    Evaluate :

   (i)  \frac{\sin 18^{o}}{\cos 72^{o}}
 

Answers (1)
M manish

\frac{\sin 18^{o}}{\cos 72^{o}}
We can write the above equation as; 
=\frac{\sin (90^0-72^0)}{\cos 72^0}
By using the identity of  \sin (90^o-\theta) = \cos \theta
Therefore, \frac{\cos 72^0}{\cos 72^0} = 1

So, the answer is 1.

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