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If cos alpha + cos beta = 1//2 and sin a...

If `cos alpha + cos beta = 1//2 and sin alpha+ sin beta = 1//3`, then

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`a to r; b to p; c to q; d to s`
`cos alpha + cos beta = 1//2`
`rArr 2cos ((alpha + beta)/( 2)) cos ((alpha - beta)/(2) ) = (1)/(2) " "` ...(i)
` sin alpha + sin beta = 1//3`
`rArr 2 sin (( alpha + beta)/(2)) cos((alpha -beta)/(2)) = (1)/(3)" "`...(ii)
Dividing Eq. (ii) by Eq. (i), we get
`" "tan((alpha + beta)/(2) ) = (2)/(3)`
`rArr cos((alpha + beta)/( 2)) = pm (3)/(sqrt(13))`
Squaring and adding the given results, we have
`2+2 cos(alpha - beta) = (13)/(36)`
or `cos (alpha -beta) = - (59)/(72)`
Now, `2cos ^(2) ((alpha -beta)/(2)) - 1 = cos (alpha -beta)`
or `2cos^(2)((alpha -beta)/(2)) = 1 - (59)/(72)= (13)/(72)`
or `cos((alpha -beta)/(2)) = pm (sqrt(13))/(12)`
or `tan((alpha -beta)/(2)) = pm sqrt((131)/(13))`
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