Evaluate the following:
$ \frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right) $

Given:

\( \frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right) \)

To do:

We have to evaluate \( \frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right) \).

Solution:  

We know that,

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

$tan\ (90^{\circ}- \theta) = cot\ \theta$

Therefore,

$\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)=\left(\frac{\cot 40^{\circ} }{\tan( 90^{\circ}-50^{\circ} )}\right) -\frac{1}{2}\left(\frac{\cos 35^{\circ} }{\sin( 90^{\circ} -35^{\circ} )}\right)$

$=\left(\frac{\cot 40^{\circ} }{\cot 40^{\circ} }\right) -\frac{1}{2}\left(\frac{\cos 35^{\circ} }{\cos 35^{\circ} }\right)$

$=1-\frac{1}{2}( 1)$

$=\frac{2( 1)-1}{2}$

$=\frac{1}{2}$

Therefore, $\frac{\cot 40^{\circ}}{\tan 50^{\circ}}-\frac{1}{2}\left(\frac{\cos 35^{\circ}}{\sin 55^{\circ}}\right)=\frac{1}{2}$.   

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Updated on: 10-Oct-2022

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