Question

Calculate the value of:
cot² (π / 14) + cot² (3π / 14) + cot² (5π / 14)

Please help me !!

Answer 1

cot²(π/14) + cot²(3π/14) + cot²(π -9π/14)
cot²(π/14) + cot²(3π/14) + cot²(9π/14)

Let ∅ = π/14

4∅ + 3∅ = π/2
4∅ = π/2 - 3∅

take tan both sides,
tan4∅ = tan(π/2 - 3∅)
tan4∅ = cot3∅
tan4∅ = 1/tan3∅
tan4∅× tan3∅ = 1
put formula ,
tan4∅ = ( 4tan∅ - 4tan³∅)/( 1-6tan²∅+tan⁴∅)

and
tan3∅ = (3tan∅ -tan³∅)/(1 - 3tan²∅)

Let tan∅ = P

(4P -4P³)/(1 -6P² + P⁴) × (3P - P³)/(1 -3P²) = 1
(4P -4P³)(3P -P³) = (1 - 6P²+P⁴)(1 - 3P²)

12P² -4P⁴ -12P⁴ + 4P^6 = 1 - 6P² + P⁴ -3P² +18P⁴ -3P^6

7P^6 -35P⁴ +21P² -1 = 0

this is cubic equation in P² . I.e P² = tan²∅
again,
put P² = 1/Q²
then, Q² = cot²∅ { we have required }

after putting P² = 1/Q²

Q^6 - 21Q⁴ + 35Q² - 7 = 0

this is cubic equation of Q² . I.e cot²∅
the roots of this equation is
cot²(π/14) , cot²(3π/14) and cot²(9π/14)

sum of roots = -b/a = 21/1 = 21

hence,

cot²(π/14) + cot²(3π/14) + cot²(9π/14) =21
so,
cot²(π/14) + cot²(3π/14) + cot²(5π/14) =21