Evaluate the following:
$ \left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2} $

Given:

\( \left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2} \)

To do:

We have to evaluate \( \left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2} \).

Solution:  

We know that,

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

Therefore,

$\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2}=\left(\frac{\sin 27^{\circ} }{\cos( 90^{\circ} -27^{\circ} )}\right)^{2} -\left(\frac{\cos( 90^{\circ} -27^{\circ} )}{\sin 27^{\circ} }\right)^{2}$

$=\left(\frac{\sin 27^{\circ}}{\sin 27^{\circ} }\right)^{2} -\left(\frac{\sin 27^{\circ} }{\sin 27^{\circ} }\right)^{2}$

$=( 1)^{2} -( 1)^{2}$

$=1-1$

$=0$

Therefore, $\left(\frac{\sin 27^{\circ}}{\cos 63^{\circ}}\right)^{2}-\left(\frac{\cos 63^{\circ}}{\sin 27^{\circ}}\right)^{2}=0$.    

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Simply Easy Learning

Updated on: 10-Oct-2022

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