Question
Mon July 30, 2012 By:

1/sin10

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Expert Reply
Mon July 30, 2012
(1/sin10) - (sq.root3/cos10)= {cos10 - ? 3 sin10}/sin10 cos10
= 2 {½ cos10 - ? 3/2 sin10}/sin10 cos10
=2(cos60cos 10- sin60sin10)/{½×2sin10 cos10} 
Since ½=cos60 and ? 3/2=sin60
2cos(60+10)/{½×sin20}=4cos70/sin20 
= 4cos70/sin(90-70) since sin20=sin(90-70)= cos70
=4cos70/cos70= 4
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