Question

Prove that root3 (cosec (20))-sec (20)=4

Answer 1

Root 3 /sin 20 - 1/cos 20

= (root3 cos 20 - sin 20 )/ (sin 20 cos 20)
(multiplying and dividing by 2)
=  2* (root3/2 cos 20 -1/2 sin 20)/ (sin 20 cos 20)

now sin 20 cos 20  = (2 sin 20 cos 20)/2 = (sin 40)/2
and writing root3/2 = sin 60, 1/2 = cos 60

= 2* (sin 60 cos 20 - cos 60 sin 20)/ (sin 40)/2

= 4* (sin (60-20))/sin 40

= 4* (sin 40)/sin 40
=4

Ask if any doubt in any step :)

Answer 2

Hello,

1) Rappels:

all angles are in degree.

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
2sin(a)cos(a)=sin(2a)
sin(a)=cos(90-a)
cos(30)=√3/2
sin(30)=1/2
2)
√3 cos(20)-sin(20)=2(√3/2 cos(20)-1/2 sin(20)
=2*(cos(30) *sin(20)-sin(30) sin(20))=2 cos(30+20)=2 cos(50)

sin(20) cos(20)=1/2 cos(40)=1/2 cos(50)

√3 (cosec (20))-sec (20) = √3 /sin(20)-1/cos(20)
=[√3*cos(20)-sin(20)]/(sin(20)cos(20))
=2 cos(50)/ (1/2cos(50))=4