1.

If `A=pi/5,`then find the value of `sum_(r=1)^8tan(r A)*tan((r+1)A)dot`

Answer» `tan((r+1)A-(rA))=(tan(r+1)A-tan(rA))/(1+tan(r+1)A*tan(rA))`
`1+tan(rA)*tan(r+1)A=(tan(r+1)A-tan(rA))/(tanA)`
`tan(rA)*tan(r+1)A=(tan(r+1)A-tan(rA))/(tanA)-1`
`S=sum_(r=1)^8(tan(r+1)A-tan(rA))/(tanA)-sum_(r=1)^8 1`
`=1/(tanA)sum_(r=1)^8(tan(r+1)A-tan(rA)-8`
`=1/(tanA)[(tan2A-tanA)+(tan3A-tan2A)+...(tan9A-tan8A]-8`
`=1/(tanA)[tan9A-tanA]-8`
`tan9A=tan(9/5pi)=tan(2pi-pi/5)`
`Ss=1/(tanA)(-tanA-tanA)-8`
`=(-2tanA)/(tanA)-8`
`=-2-8`
`S=-10`.


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