If `cos alpha+cos beta=0=sin alpha+sinbeta, then cos2alpha+cos 2beta=`

Options

A. `-2sin (alpha+beta)`
B. `-2cos(alpha+beta)`
C. `2sin(alpha+beta)`
D. `2cos(alpha+beta)`

Answer

B

Explanation

`(cos alpha+cos beta)^(2)-(sin alpha+sinbeta)^(2)=0`
or `(cos^(2)alpha+cos^(2)beta+2cos alpha cos beta)^(-)`
`(sin^(2)alpha+sin^(2)beta+2sin alpha sin beta)`
or `cos 2alpha+cos 2beta=-2(cos alpha cos beta-sin alpha sin beta)`
`=-2cos(alpha+beta)`.