Question Number 101554 by student work last updated on 03/Jul/20

(1/(cos80))−((√3)/(sin80))=?

Answered by bramlex last updated on 03/Jul/20

((sin 80^o −(√3) cos 80^o )/(sin 80^o cos 80^o )) = ((2(sin 80^o −(√3) cos 80^o ))/(sin 160^o )) = 4× (((1/2)sin 80^o −((√3)/2)cos 80^o )/(sin 20^o )) = ((sin 30^o sin 80^o −cos 30^o cos 80^o )/(sin 20^o )) = ((−4{cos 110^o })/(sin 20^o )) = ((−4(cos (90^o +20^o )))/(sin 20^o )) = 4 ★

Answered by Dwaipayan Shikari last updated on 03/Jul/20

((sin80°−(√3)sin80°)/((1/2)sin20°))=4((((sin80°)/2)−((√3)/2)cos80°)/(sin20))=4((cos(−110°))/(sin20))=4★■

Answered by mahdi last updated on 03/Jul/20

=((sin80−(√3)cos80)/(cos80sin80))=4(((1/2)sin80−((√3)/2)cos80)/(2cos80sin80)) =4((cos60sin80−sin60cos80)/(2cos80sin80))= 4((sin(80−60))/(sin(80+80)))=4((sin20)/(sin160))=4((sin20)/(sin20))=4