4cos20 minus root 3 cot 20
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Given, √3*cot 20 - 4*cos 20
= √3*cos 20/sin 20 - 4*cos 20
= (√3*cos 20 - 4*cos 20*sin 20)//sin 20
= (√3*cos 20 - 2*2*cos 20*sin 20)//sin 20
= {√3*cos 20 - 2*sin (2*20)}//sin 20
= {√3*cos 20 - 2*sin 40}//sin 20 {since sin 2x = 2*sin x*cosx}
= {2*sin 60*cos 20 - 2*sin 40}//sin 20 {since sin 60 = √3/2}
= {sin 80 + sin 40 - 2*sin 40}//sin 20 {since 2*sin A*cos B = sin (A + B) + sin (A - B)}
= {sin 80 - sin 40}//sin 20
= {2*cos 60*sin 20}/sin 20 {since sin A - sin B = 2 * sin (A - B)/2 *cos (A + B)/2}
= 2*cos 60
= 2 * 1/2 {since cos 60 = 1/2]
= 1
So, √3*cot 20 - 4*cos 20 = 1
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