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tan square pi by 16+ tan square 2pi by 16+tan square 3pi by 16+tan square 4pi by 16+ tan square 5pi by 16+ tan square 6pi by 16+tan square 7pi by 16=35

tan square pi by 16+ tan square 2pi by 16+tan square 3pi by 16+tan square 4pi by 16+ tan square 5pi by 16+ tan square 6pi by 16+tan square 7pi by 16=35

Grade:11

1 Answers

Arun
25763 Points
4 years ago
we use formula tan^2a=(1-cos2a)/(1+cos2a)
tan^(pi/16)+tan^2(7pi/16)+tan^(2pi/16)+tan^2(6pi/16)+tan^2(3pi/16)+tan^2(5pi/16)+tan^(4pi/16)
tan^(pi/16)+tan^2(pi/2-pi/16)+tan^2(2pi/16)+tan^2(pi/2-2pi/16)+tan^2(3pi/16)+tan^2(pi/2-3pi/16)+tan^2(pi/4)
tan^2(pi/16)+cot^2(pi/16)+tan^2(2pi/16)+cot^2(2pi/16)+tan^2(3pi/16)+cot^2(3pi/16)+1
we use formula
tan^2a+cot^2a=(1-cos2a)/(1+cos2a)+(1+cos2a)/(1-cos2a)
=2(1+cos^2(2a))/sin^2(2a))
2(1+cos^2(pi/8))/sin^2(pi/8) +2(1+cos^2(pi/4))/sin^2(pi/4) +2(1+cos^2(3pi/8))/sin^2(3pi/8) +1
2(cosec^2(pi/8)+cot^2(pi/8)+cosec^2(3pi/8)+cot^2(3pi/8)+2(1+1/2)/(1/2) +1
2(1+2cot^2(pi/8)+1+2cot^2(3pi/8))+6+1
4+6+1+4(cot^2(pi/8)+cot^2(3pi/8))
11+4{(1+cospi/4)/(1-cos(pi/4)) + (1+cos(3pi/4))/(1-cos(3pi/4)}
11+4{(1+1/sqrt2)/(1-1/sqrt2) +(1-1/sqrt2)/(1+1/sqrt2)}
11+4{2*3/2)/(1/2)
11+4(6)
11+24
35

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