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The value of sec1(1410k=0sec(7π12+kπ2)sec(7π12+(k+1)π2)) in the interval [π4,3π4] equals

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Solution

sec1⎜ ⎜1410k=01cos(7π12+kπ2)cos(7π12+(k+1)π2)⎟ ⎟
=sec1⎜ ⎜1410k=0sin[(7π12+(k+1)π2) (7π12+kπ2)]cos(7π12+kπ2)cos(7π12+(k+1)π2)⎟ ⎟
(sin(AB)=sinAcosBcosAsinB)
=sec1⎜ ⎜1410k=0(sin(7π12+(k+1)π2)cos(7π12+kπ2))(cos(7π12+(k+1)π2)sin(7π12+kπ2))cos(7π12+kπ2)cos(7π12+(k+1)π2)⎟ ⎟
=sec1(1410k=0(tan(7π12+(k+1)π2))(tan(7π12+kπ2)))
=sec114(tan(7π12+π2)tan(7π12)+tan(7π12+2π2)tan(7π12+π2)+ ....+tan(7π12+11π2)tan(7π12+10π2))
=sec114(tan(7π12+11π2)tan7π12)
=sec114(tan(6π+π12)tan(π5π12))
=sec114(tanπ12+tan5π12)
=sec114[(23)+(2+3)]
=sec1(1)=0

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