2010/011) If cos(x)/cos(y) = a/b, then a tan(x) + b tan(y) equals

A) (a+b)cot((x+y)/2); B) (a+b)tan((x+y)/2); ...?
C) (a+b)(tan(x/2) + tan(y/2));
D) (a+b)(cot(x/2) + cot(y/2));
Kindly explain your answer...

we have (a+b) on RHS of (a)(b)(c) and (d) so let us eliminate a and b from numerator of
( a tan(x) + b tan(y))/(a+b)

b cos x = a cos y
b / a = cos y/ cos x
(b+a)/a = (cos y + cos x)/cos x

a/(a+b) = cos x/(cos y+ cos x)
and b/(a+b) = cos y /(cos y + cos x)

(a tan x + b tan y)/(a+b)
= (a/(a+b)) tan x + (b/(a+b) tan y
= (cos x tan x + cos y tan y)/(cos y + cos x)
= (sin x + sin y)/(cos x + cos y)
= (2 sin (x+y)/2 cos (x-y)/2 )/(2 cos (x+y)/2 cos (x-y)/2 )
= tan (x+y)/2

hence ans is B