to prove(r=1 to r=7)∑tan2 rπ/16=35

ANS:

Let,

a=∏/16

tan7a=cota             since (∏/2-7∏/16=∏/16)

similarly we get,

tan6a=cot2a

tan5a=cot3a

and 4a =∏/4  tan4a=1

therefore given expression reduces to,

tan^2a+cot^2a+tan^2(2a)+cot^2(2a)+tan^2(3a)+cot^2(3a)+1

tan^2a+cot^2a=

=sin^4a+cos^4a/sin^2acos^2a

=((sin^2a+cos^2a)^2 – 2sin^2acos^2a)/ sin^2acos^2a

=(1– 2sin^2acos^2a) / sin^2acos^2a

=1/ sin^2acos^2a -2

=(4/4 sin^2acos^2a )-2

=4/sin^2(2a) -2

In the same way simplifying other sums we get,

4(1/sin^2(2a)+1/sin^2(4a)+1/sin^2(6a))-5

1/sin^2(4a)=2

=4(1/sin^2(2a)+1/sin^2(6a))+3

=8(1/(1-cos4a)+1/(1-cos12a))+3

[4a=∏/4  and  12a =3∏/4]

this gives

=32+3

=35