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If cosalpha+cosbeta=0=sinalpha+sinbeta, ...

If `cosalpha+cosbeta=0=sinalpha+sinbeta,` then `cos2alpha+cos2beta` is equal to `-2"sin"(alpha+beta)` (b) `-2cos(alpha+beta)` `2"sin"(alpha+beta)` (d) `2"cos"(alpha+beta)`

A

`-2sin(alpha+beta)`

B

`-2cos(alpha+beta)`

C

`2sin(alpha+beta)`

D

`2cos(alpha+beta)`

Text Solution

Verified by Experts

The correct Answer is:
C
`cos alpha + cos beta = 0 = sin alpha + sin beta`
Squaring and adding, we get
`2+2 cos (alpha-beta)=0`
`therefore cos (alpha - beta)=-1`
`therefore cos 2alpha + cos 2beta`
`=2 cos (alpha+beta)cos (alpha-beta)`
`=-2 cos (alpha+beta)`
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