If `cos alpha+cos beta=3/2 and sin alpha+sin beta=1/2 and theta` is the arithmetic mean of `alpha and beta,` then `sin 2theta+cos 2theta` is equal to :

Options

A. `(3)/(5)`
B. `(7)/(5)`
C. `(4)/(5)`
D. `(8)/(5)`

Answer

B

Explanation

`cosalpha+cosbeta=(3)/(2)andsinalpha+sinbeta=(1)/(2)`
`rArr2cos((alpha+beta)/(2))cos((alpha+beta)/(2))=(3)/(2)`
and `2sin((alpha+beta)/(2))cos((alpha-beta)/(2))=(1)/(2)`
`rArrtan((alpha+beta)/(2))=(1)/(3)`
`becausetheta=(alpha+beta)/(2)` [Given]
`rArr2theta=alpha+beta`
`thereforesin2theta+cos2theta=sin(alpha=beta)+cos(alpha+beta)`
`=(2tan((alpha+beta)/(2)))/(1+tan^(2)((alpha+beta)/(2)))+(1-tan^(2)((alpha+beta)/(2)))/(1+tan^(2)((alpha+beta)/(2)))`
`[becausesin2theta=(2tantheta)/(1-tan^(2)theta),cos2theta=(1-tan^(2)theta)/(1+tan^(2)theta)]`
`=(2((1)/(3)))/(1+((1)/(3))^(2))+(1-((1)/(3))^(2))/(1+((1)/(3))^(2))=(2)/(3)xx(9)/(10)+(8)/(9)xx(9)/(10)=(6)/(10)+(8)/(10)=(7)/(5)`