Evaluate each of the following:$ \frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ} $

Given:

\( \frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ} \)

To do:

We have to evaluate \( \frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ} \).

Solution:  

We know that,

$cot 30^{\circ}=\sqrt3$

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

$\cos 45^{\circ}=\frac{1}{\sqrt2}$

$\frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ}=\frac{4}{\left(\sqrt{3}\right)^{2}} +\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^{2}} -\left(\frac{1}{\sqrt{2}}\right)^{2}$

$=\frac{4}{3} +\frac{1}{\frac{3}{4}} -\frac{1}{2}$

$=\frac{4}{3} +\frac{4}{3} -\frac{1}{2}$

$=\frac{4( 2) +4( 2) -1( 3)}{6}$

$=\frac{8+8-3}{6}$

$=\frac{13}{6}$

Hence, $\frac{4}{\cot ^{2} 30^{\circ}}+\frac{1}{\sin ^{2} 60^{\circ}}-\cos ^{2} 45^{\circ}=\frac{13}{6}$. 

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Updated on: 10-Oct-2022

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