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, $\left(\frac{\sin 49^{\circ}}{\cos 41^{\circ}}\right)^{2}+\left(\frac{\cos 41^{\circ}}{\sin 49^{\circ}}\right)^{2}=2$.   

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Updated on: 10-Oct-2022

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