If `x+y=pi//4` and `tan x+tan y=1`, then `(n in Z)`

Options

A. `sin x=0` always
B. when `x=npi+pi//4` then `y=-npi`
C. when `x=npi` then `y=npi +(pi//4)`
D. when `x=npi+pi//4` then `y=npi -(pi//4)`

Answer

B, C

Explanation

From `tanx+tan y=1`, we have
`(sin x)/(cos x)+(sin y)/(cos y)=1`
or `sin x cos y+ sin y cos x= cos x cos y`
or `2 sin (x+y) =2 cos x cos y`
or `2 sin (x+y) = cos (x+y) + cos (x-y)`
or `2 sin (pi/4) = cos (pi/4) + cos (c-y)`
or `cos (x-y)=1/sqrt(2)= cos (pi/4)`
`rArr x-y=2n pi pm (pi//4), AA n in Z` ...(i)
Also we have `x+y= pi//4` ...(ii)
From Eqs. (i) and (ii), we have
`x= n pi + (pi//4) and y= -n pi, AA n in Z`
or `x = npi and y=- n pi + pi//4, AA n in Z`