Saved Bookmarks
| 1. |
The value of `sec^(-1)((1)/(4)sum_(k=0)^(10)sec((7pi)/(12)+(kpi)/(2))sec((7pi)/(12)+((k+1)pi)/(2)))` in the inerval `[-(pi)/(4),(3pi)/(4)]` equals.......... |
|
Answer» `therefore =sum_(k=0)^(10)sec((7pi)/(12)+(kpi)/(2))sec((7pi)/(12)+((k+1)pi)/(2))` `=Sigma_(k=0)^(10) (1)/(cos ((7pi)/(12)+(kpi)/(2))cos((7pi)/(12)+((k+1)pi)/(2)))` ` =Sigma_(k=1)^(10) (sin [((7pi)/(12)+((k+1)pi)/(2))-((7pi)/(12)+(kpi)/(2))])/(cos ((7pi)/(12)+(kpi)/(2))cos ((7pi)/(12)+((k+1)pi)/(2)))` `[because (7pi)/(12)+((k+1)pi)/(2)-((7pi)/(12)+(kpi)/(2))=(pi)/(2) and sin (pi)/(2)=1]` `sin ((7pi)/(12)+((k+1)pi)/(2))cos ((7pi)/(12)+(kpi)/(2))` `=Sigma_(k=0)^(10) (-sin ((7pi)/(12)+(kpi)/(2))cos ((7pi)/(12)+((k+1)pi)/2))/(cos ((7pi)/(12)+(kpi)/(2))cos ((7pi)/(12)+((k+1)pi)/(2)))` `= Sigma _(k=0)^(10) [tan ((7pi)/(12)+((k+1)pi)/(2))-tan ((7pi)/(12)+(kpi)/(2))]` `=tan ((7pi)/(12)+(pi)/(2))-tan ((7pi)/(12))+tan ((7pi)/(12)+(2pi)/(2))-tan ((7pi)/(12)+(pi)/(2))+tan ((7pi)/(12)+(11pi)/(2))-tan((7pi)/(12)+(pi)/(2))` `=tan ((7pi)/(12)+(11pi)/(2))-tan (7pi)/(12)=tan(pi)/(12)+cot (pi)/(12)` `=(1)/(sin(pi)/(12)cos. (pi)/(12))=(2)/(sin.(pi)/(6))=4` So, `sec^-1 ((1)/(4)Sigma_(k=0)^(10)sec((7pi)/(12)+(kpi)/(2))sec((7pi)/(12)+((k+1)pi)/(2)))=sec^-1(1)=0` |
|