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Evaluate the following:
$ \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2} $
Given:
\( \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2} \)
To do:
We have to evaluate \( \left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2} \).
Solution:
We know that,
$sin\ (90^{\circ}- \theta) = cos\ \theta$
Therefore,
$\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}=\left(\frac{\sin( 90^{\circ} -41^{\circ} )}{\cos 41^{\circ} }\right)^{2} +\left(\frac{\cos 41^{\circ} }{\sin( 90^{\circ} -41^{\circ} )}\right)^{2}$
$=\left(\frac{\cos 41^{\circ}}{\cos 41^{\circ} }\right)^{2} +\left(\frac{\cos 41^{\circ} }{\cos 41^{\circ} }\right)^{2}$
$=( 1)^{2} +( 1)^{2}$
$=1+1$
$=2$
Therefore, $\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}=2$.