If `cos(x/2) cos (x/2^2) cos (x/2^) ..... cos (x/2^n)=sinx/(sin(x/2^n)` prove that` 1/2tan(x/2)+1/4tan(x/4).....1/2^(2n)tan(x/2^n)=1/2^ncot(x/2^n)-cotx`

Options

A. `(f'(x))/(f(x))`
B. `(f(x))/(f'(x))`
C. `(-f'(x))/(f(x))`
D. 0

Answer

C

Explanation

`f(x) = cos ""(x)/(2) "cos" (x)/(2^(2))"cos" (x)/(x_(3)) ... " cos "(x)/(2^(n))`
Taking log on both sides , we get
`log f(x) = logcos ""(x)/(2)+ "log cos" (x)/(2^(2))+ ... "log cos "(x)/(2^(n))`
On differentiating both sides w.r.t.w, we get
`(f'(x))/(f(x) )=- ((1)/(2) "tan" (x)/(2) + (1)/(2^(2)) "tan" ((x)/(2^(2))) + ...+ (1)/(2^(n)) "tan" (x)/(2^(n)))`
`rArr (f'(x))/(f(x) )=- (1)/(2) "tan" (x)/(2) + (1)/(2^(2)) "tan" (x)/(2^(2)) + ...+ (1)/(2^(n)) "tan" (x)/(2^(n))`