Question

root 3 cot 20 - 4 cos 20

Answer 1

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

Answer 2

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