Question

Evaluate each of the following $\frac{tan45^{0}}{cosec30^{0}}+\frac{sec60^{0}}{cot45^{0}}-\frac{5sin90^{0}}{2cos0^{0}}$

Answer 1

SOLUTION :  

Given :  

tan45° / cosec 30° + sec 60° / cot 45° -  5 sin 90° / 2 cos 0°

= ½ + 2/1 – 5(1)/2(1)

[tan 45° = 1, cosec 30° =  sec 60°= 2,  cot 45° = 1, sin 90° = 1 ,  cos 0°= 1]

= ½ + 2/1 - 5/2

=( ½ - 5/2) + 2/1

[By rearranging the terms]

= (1 - 5)/2 + 2/1

= - 4/2 +2  

= -2 + 2  = 0

tan45° / cosec 30° + sec 60° / cot 45° -  5 sin 90° / 2 cos 0° = 0

Hence, tan45° / cosec 30° + sec 60° / cot 45° -  5 sin 90° / 2 cos 0° = 0

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

Hi there!
Here's the answer:

•°•°•°•°•°<><><<><>><><>•°•°•°•°•°

¶¶¶ We have

• Tan 45° = 1
• Cosec 30° = 2
• Sec 60° = 2
• Cot 45° = 1
• Sin 90° = 1
• Cos 0° = 1

•°•°•°•°•°<><><<><>><><>•°•°•°•°•°

Given,

$\frac{tan45^{0}}{cosec30^{0}}+\frac{sec60^{0}}{cot45^{0}}-\frac{5sin90^{0}}{2cos0^{0}}$

Substituting Values

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

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

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

= -2 + 2

= 0

•°•°•°•°•°<><><<><>><><>•°•°•°•°•°

©#£€®$

:)

Hope it helps