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,

$tan 45^{\circ}=1$

$cosec 30^{\circ}=2$

$\sec 60^{\circ}=2$

$\cot 45^{\circ}=1$

$\sin 90^{\circ}=1$

$\cos 0^{\circ}=1$

$\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}}=\frac{1}{2} +\frac{2}{1} -\frac{5( 1)}{2( 1)}$

$=\frac{1}{2} +2-\frac{5}{2}$

$=\frac{1+2( 2) -5}{2}$

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

$=0$

Hence, $\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}}=0$.   

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Updated on: 10-Oct-2022

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