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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
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