If `sin x + sin y = a and cos x + cos y=b,` then `tan^2\ (x+y)/2 + tan^2\ (x-y)/2` is equal to

Options

A. `(a^(4)+b^(4)+4b^(2))/(a^(2)b^(2)+b^(4))`
B. `(a^(4)-b^(4)+4b^(2))/(a^(2)b^(2)+b^(4))`
C. `(a^(4)-b^(4)+4b^(2))/(a^(2)b^(2)+a^(4))`
D. None of the above

Answer

B

Explanation

sinx + sin y = a
`rArr2sin((x+y)/(2))cos((x-y)/(2))=a` ... (1)
cos x + cosy = b
`rArr2cos((x+y)/(2))cos((x-y)/(2))=b` . . . (2)
dividing (1) & (2)
`tan((x+y)/(2))=(a)/(b)`
Squaring eq (1) & (2) and adding-
`4cos^(2)((x-y)/(2))=a^(2)+b^(2)`
`sec^(2)((x-y)/(2))=(4)/(a^(2)+b^(2))`
again , `tan^(2)((x+y)/(2))+tan^(2)((x-y)/(2))`
`=((a)/(b))^(2)+sec^(2)((x-y)/(2))-1`
`=(a^(2))/(b^(2))+(4)/(a^(2)+b^(2))-1=(a^(4)-b^(4)+4b^(2))/(a^(2)b^(2)+b^(4))`