Question

2sin²30 cot30 – 3cos²60 sec²30.Evaluate it.

Answer 1

HELLO DEAR,

We know:- cotx = cosx/sinx
secx = cosec(90 - x) , sin(90 - x) = cosx

Now, 2sin²30*cot30 - 3cos²60*sec²30

= 2sin²30 * cos30/sin30 - 3cos²60*sec²(90 - 60)

= 2sin30*cos30- 3cos²60*cosec²60
[2sinxcosx = sin2x]

= sin2(30) - 3cos²60/sin²60

= sin60 - 3cot²60

= √3/2 - 3(1/√3)²

= √3/2 - 1

= (√3 - 2)/2.


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Given : 2sin²30° cot 30° – 3cos²60 sec²30

= 2(sin 30°)² cot 30° – 3 (cos 60°)² (sec30°)²

= 2 × (½)² × √3 - 3 (½)² × (2/√3)²

[sin 30° = ½, cot 30° = √3, cos 60°= ½, sec 30 = 2/√3]

= 2×1/4 × √3 - 3×1/4 × 4/3

= √3/2 - 1

= (√3- 2)/2

HOPE THIS ANSWER WILL HELP YOU...