In `DeltaABC`, with usual notations, if a,b,c are ni AP then a `cos^(2)((C)/(2))+c" "cos^(2)((A)/(2))=`

Options

A. `(3a)/(2)`
B. `(3c)/(2)`
C. `(3b)/(2)`
D. `(3abc)/(2)`

Answer

C

Explanation

Given a,b,c are in AP
`therefore2b=a+c`
Now, `acos^(2)(C)/(2)+c " cos"^(2)(A)/(2)`
`implies a((1+cosC)/(2))+c((1+cosA)/(2))" "[because2"cos"^(2)(x)/(2)=1+cosx]`
`=(1)/(2)[a+c+a" cos C"+c" cos A"]`
`=(1)/(2)[2b+b]=(3b)/(2)" "[becauseb=acosC+c" "cos" "A]`