Question

Prove that cot ($\frac{\pi}{20}$) . cot ($\frac{3\pi}{20}$) . cot ($\frac{5\pi}{20}$) . cot ($\frac{7\pi}{20}$) . cot ($\frac{9\pi}{20}$) = 1

Answer 1

LHS = cot(π/20) cot(3π/20) . cot(5π/20) . cot(7π/20) . cot(9π/20)

= cot(π/2 - 9π/20) . cot(π/2 - 7π/20) . cot(5π/20) . cot(7π/20) . cot(9π/20)

we know, cot(π/2 - x) = tanx

so, cot(π/2 - 9π/20) = tan(9π/20)
cot(π/2 - 7π/20) = tan(7π/20)

= tan(9π/20). tan(7π/20) . cot(5π/20) . cot(7π/20) . cot(9π/20)

= {(tan(9π/20).cot(9π/20)} . {tan(7π/20).cot(7π/20)} cot(π/4)

we know, tanx.cotx = 1
so, (tan(9π/20).cot(9π/20)= 1
tan(7π/20).cot(7π/20)} = 1

= 1 . cot(π/4) . 1 [ as cot(5π/20) =cot(π/4) ]

= 1 . 1 . 1 = 1 = RHS

Answer 2

HELLO DEAR,



we know,
cot(π/2 - x) = tanx
tanx.cotx = 1
cot(5π/20) =cot(π/4)

now,
cot(π/20) cot(3π/20) * cot(5π/20) * cot(7π/20) * cot(9π/20)

=> cot(π/2 - 9π/20) * cot(π/2 - 7π/20) * cot(5π/20) * cot(7π/20) * cot(9π/20)


=> tan(9π/20) * tan(7π/20) * cot(5π/20) * cot(7π/20) * cot(9π/20)

=> {tan(9π/20) * cot(9π/20)} * {tan(7π/20) * cot(7π/20)} cot(π/4)

=> 1 × cot(π/4) × 1

= 1 × 1 × 1 = 1



I HOPE IT'S HELP YOU DEAR,
THANKS