Question

"Question15
Evaluate, tan²60° + 4cos²45°+ 3cosec²60°+2cos²90° / 2cosec 30° + 3sec60° - 7/3 cot²30°
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 288"

Answer 1

Hey there!

We know that,
tan60 = √3
cos45 = 1/√2
cosec60 = 2/√3
cos90 = 0
cosec30 = 2
sec60 = 2
cot30 = √3

Now,
tan²60° + 4cos²45°+ 3cosec²60°+2cos²90° / [ 2cosec 30° + 3sec60° - 7/3 cot²30° ]

= [ (√3)² + 4 ( 1/√2)² + 3(2/√3)² + 2( 0²) ] / [ 2(2)+ 3(2) -7/3 ( √3)² ]

= [ 3 + 4(1/2 ) + 3(4/3) ] [ 4 + 6 - 7/3 ( 3 ) ]

= [ 3 + 2 + 4 ] [ 10 - 7 ]

= 9/3

= 3 .

Therefore, tan²60° + 4cos²45°+ 3cosec²60°+2cos²90° / (2cosec 30° + 3sec60° - 7/3 cot²30° ) = 3

Answer 2

HELLO DEAR,

tan²60° + 4cos²45° + 3cosec²60° + 2cos²90°/2cosec30° + 3sec60° - 7/3cot²30°

= [(√3)² + 4(1/√2)² + 3(2/√3)² + 2(0)²] / [2(2) + 3(2) - 7/3(√3)² ]

= [ 3 + 2 + 2 + 2 + 0] / [ 4 + 6 - 7 ]

=> 9/3

=> 3

I HOPE ITS HELP YOU DEAR,
THANKS