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If `x+y=pi//4` and `tan x+tan y=1`, then `(n in Z)`A. `sin x=0` alwaysB. 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)` |
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Answer» Correct Answer - B::C 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` |
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