Question

cot pi/16*cot 2pi/ 16*cot 3pi/16.....cot 7pi/16​ ​

Answer 1

cot(π/16) . cot(2π/16) . cot(3π/16) .... cot(7π/16)

we know, cot(π/2 - θ) = tanθ

so, cot(7π/16) = cot(π/2 - π/16) = tan(π/16)

similarly, cot(6π/16) = cot(π/2 - 2π/16) = cot(2π/16)

cot(5π/16) = cot(π/2 - 3π/16) = cot(3π/16)

now, cot(π/16) . cot(2π/16) . cot(3π/16) . cot(4π/16) . tan(3π/16) . tan(2π/16) . tan(π/16)

= {tan(π/16). cot(π/16)}. {tan(2π/16). cot(2π/16)} . {tan(3π/16). cot(3π/16)}. cot(4π/16)

we know, tanθ. cotθ = 1

= 1 × 1 × 1 × cot(4π/16)

= cot(π/4)

= 1 [Ans]

Answer 2

Answer:

cot(π/16) . cot(2π/16) . cot(3π/16) .... cot(7π/16)

we know, cot(π/2 - θ) = tanθ

so, cot(7π/16) = cot(π/2 - π/16) = tan(π/16)

similarly, cot(6π/16) = cot(π/2 - 2π/16) = cot(2π/16)

cot(5π/16) = cot(π/2 - 3π/16) = cot(3π/16)

now, cot(π/16) . cot(2π/16) . cot(3π/16) . cot(4π/16) . tan(3π/16) . tan(2π/16) . tan(π/16)

= {tan(π/16). cot(π/16)}. {tan(2π/16). cot(2π/16)} . {tan(3π/16). cot(3π/16)}. cot(4π/16)

we know, tanθ. cotθ = 1

= 1 × 1 × 1 × cot(4π/16)

= cot(π/4)

= 1/$\sqrt{2}$[Ans]