Question

pleaseans trignometry class 12th​

Answer 1

ur question

[{singma} from r=1 to r=8]tan(rA).tan(r+1)A,where A=36°

Solution:

=>Let us first consider,the value of Tan{(r+1)A-tanrA}

Now,using the identity Tan(A-B)=(TanA-TanB)/1+TanA.TanB,,we get

=>Tan{(r+1)A-rA}={Tan(r+1)A-TanrA}/{1+tan(r+1)A.tanrA)}......

=>Tan{rA+A-rA}={Tan(r+1)A-TanrA}/{1+tan(r+1)A.TanrA}

=>Tan(r+1)A.TanrA={Tan(r+1)A-TanrA}/TanA -1}

Now, applying,singma on both side we get we get

[{singma} from r=1 to r=8]tan(rA).tan(r+1)A,=£from r=1 to 8{Tan(r+1)A-TanrA}/TanA -1}

=>1/TanA[sigma from 1 to 8]tan(r+1)A-tanrA}-{sigma 1 to 8}1

=>put r=1 to 8

=>1/TanA{tan2A-TanA+Tan3A-Tan2A+Tan4A-Tan3A+Tan5A-tan4A+Tan6A-tan5A+tan7A-tan6A+tan7A-tan8A+tan9A-tan8A}-8

now,we can clearly see

every terms ,will be cancelled out ,we get

=> 1/tanA×(Tan9A-TanA)-8

now,

now,we know

A=36=π/5

Tan9A=tan(2π-π/5)

this will lie in the fourth coordinates

so,

tan(2π-π/5)=-tanπ/5=-tanA

putting this value of tan9A we get

=>1/tanA(-tanA-tanA-8)

=>(-2tanA)/tanA-8

=>-2-8=-10

{hope it helps you}