# sinx/siny=1/2,  cosx/cosy=3/2, when x,y belong to [theta,pi/2] then find value of tan (x+y)

Ravi
9 years ago
Expand the tan(x+y) to simple form and substitute the value of tanx(tan y) from the relation obtained by dividing thesinx/siny=1/2, cosx/cosy=3/2,

Next, find the value of either tan x (or tan y), by substituting in the identity the value obtained fromsinx/siny=1/2, cosx/cosy=3/2. Solve.
Sai Ram
13 Points
5 years ago
sin x/sin y=1/2 , cosx/cosy = 3/2 then tan(x+y)
tan x/ tany=sinx/cosx/siny/cosy
=1/2/3/2
=1/3
3tanx=tany
tan(x+y)=tanx+3tanx/1-tanx.3tanx
=4tanx/1-3tanx2   _______(eq--1)
2sinx=siny  cosy=2/3x
w.k.t sin2y+cos2y=1
==>  (2sinx)2+(2/3 cosx)2 = 1
==> 4 sin2x + 4/9 cos2x    =1
==> 4cos2x(sin2x/cos2x + 1/9 )=1
==>.   tan2x +1/9 = 1 / 4cos2
==>   (9tan2x + 1)/9 = 1/4 sec2x
==>  36 tan2x + 9 = 9sec2x
==>  36 tan2x + 9 = 9(1+tan2x)
Solve  the equation

CHANCHAL GUPTA
13 Points
5 years ago
Sin x/sin y=1/2. Cos x/cos y=3/2
So tan x/tan y=1/3 and sec x/sec y =2/3
tan 2 y =9 tan 2x and sec 2y=9/4 sec 2x
sec 2y-tan 2y=1
9/4 sec 2x -9 tan 2 x = 1
9sec 2 x - 36 tan 2x=4
9+9 tan 2x - 36 tan 2x=4
27tan 2x=5
tan 2x = 5/27
So, tan 2y=5/3
tan(x+y)=tan x + tan y/1-tan x tan y
tan(x+y)=√5/3√3+√5/√3  /  1-√5/3√3×√5/√3
=√15