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Evaluate each of the following:$ \frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}} $
Given:
\( \frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}} \)
To do:
We have to evaluate \( \frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}} \).
Solution:
We know that,
$tan 45^{\circ}=1$
$cosec 30^{\circ}=2$
$\sec 60^{\circ}=2$
$\cot 45^{\circ}=1$
$\sin 90^{\circ}=1$
$\cos 0^{\circ}=1$
Therefore,$\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}}=\frac{1}{2} +\frac{2}{1} -\frac{5( 1)}{2( 1)}$
$=\frac{1}{2} +2-\frac{5}{2}$
$=\frac{1+2( 2) -5}{2}$
$=\frac{-4+4}{2}$
$=0$
Hence, $\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}}=0$.