The sum of all the solution in `[0,4pi]` of the equation `tanx+cotx+1=cos(x+pi/4)i s` `3pi` (b) `pi/2` (c) `(7pi)/2` (d) `4pi`

Options

A. `3 pi`
B. `pi//2`
C. `7pi//2`
D. `4pi`

Answer

C

Explanation

`tan x + cot x +1 = cos (x+pi/4)`
or `tan x+cot x=cos(x+pi/4)-1`
Now `tan x +cot x le-2` and `cos(x+pi/4)-1 ge -2`
It implies that equality holds when both are `-2`. Thus,
`cos(x+pi/4)=-1`
`rArr x+pi/4=(2m+1)pi, m in Z`
`rArr x=(3pi)/4` or `(11pi)/4`
Therefore, the sum of the solutions is
`(3pi)/4+(11pi)/4=(7pi)/2`