Question

if 1+cos(x-y)=0 then a)cosx-cosy=0 b)cosx+cosy=0 c)cosx-siny=0 d) sinx+cosy=0​

Answer 1

Answer:

option (b) is correct

Step-by-step explanation:

$\text{Given:}$

$1+cos(x-y)=0$

$cos(x-y)=-1$

$\implies\,x-y=\pi$

$\implies\,x=y+\pi$

$\implies\,cosx=cos(y+\pi)$

$\implies\,cosx=cos(\pi+y)$

$\boxed{\bf\,cos(\pi+A)=-cosA}$

$\implies\,cosx=-cosy$

$\implies\boxed{\bf\,cosx+cosy=0}$

Answer 2

cos (x-y) = -1

x-y = pi

X= pi+y

apply cos

cosx = cos ( pi+ y)

cosx = -cosy

cosx +cosy =0

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