# Evaluate each of the following

Question:

Evaluate each of the following

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

Solution:

We have,

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

Now,

$\sin 90^{\circ}=\cos 0^{\circ}=1, \tan 45^{\circ}=\cot 45^{\circ}=1, \operatorname{cosec} 30^{\circ}=\sec 60^{\circ}=2$

We get,

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

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

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

Now by taking terms with denominator 2 together and solving

We get,

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

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

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

Now $\frac{-4}{2}$ gets reduced to $-2$

Therefore,

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

$=-2+2$

$=0$

Therefore,

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