Question

Prove that 1 /2 tan ( x /2)+ 1 /4tan( x /4)+...+( 1 /2 n )tan ( x /2 n )=( 1 /2 n )cot( x /2 n ) - cotx for all n (- N and 0

Answer 1

$\frac{1}{2}tan\frac{x}{2}+\frac{1}{2^2}tan{x}{2^2}+....+\frac{1}{2^n}tan\frac{x}{2^n}=\frac{1}{2^n}Cot\frac{x}{2^n}-Cot\ x,\ \ \ \forall\ n\ \textgreater \ 0$

we can prove this by using the principle of mathematical induction.

for n = 1,  
  LHS = 1/2 * tan (x/2)
  RHS = 1/2 Cot(x/2)  - Cot x
           = 1/2 Cot(x/2) - (1- Tan^2 x/2)/(2 Tan x/2)
           = (1 - 1 + tan^2 x/2)/(2 tan x/2) = LHS

Let us assume that given identity is true for n.   We will use the formula for expansion of Cot (2A) = 1/tan (2A) like in the above for Cotx.   For n+1,

LHS = [1/2^n * Cot (x/2^n) - Cot x ] + 1/2^(n+1) * Tan (x/2^(n+1))
  = - Cot x +[1- tan^2 (x/2^(n+1)) -Tan^2 (x/2^(n+1))]
                               / tan(x/2^(n+1) / 2^(n+1)
        = - Cot x + 1/2^(n+1) * Cot (x/2^(n+1))
        = RHS  for  n+1

Hence proved by MI

Answer 2

Step-by-step explanation:

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