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Madhukar Gupta

· started a discussion

· 1 Months ago

short soln (Question: 110258.0 )

let x 60 y 30 then a=3b
now tanx/tany= 3
from option 1 a/b= 3b/b=3

Question:
If \(\cfrac{sin(x+y)}{sin(x-y)}\ =\cfrac{a+b}{a-b}\) , then the value of \(\cfrac{tan\ x}{tan\ y}\) is:
Options:
A) \(\cfrac{a}{b}\)
B) \(\cfrac{b}{a}\)
C) ab
D) \(\cfrac{a-b}{a+b}\)
Solution:
Ans: (a) 

\(\cfrac{sin(x+y)}{sin(x-y)}=\cfrac{a+b}{a-b}\)        .............(i)

\(\therefore \cfrac{sin (x+y)+sin(x-y)}{sin(x+y)-sin(x-y)}\ \ \ \ \ =\cfrac{a+b+a-b}{a+b-a+b}\)

 \(\therefore \cfrac{sin \ x\ cos\ y + cos \ x \ sin \ y + sin \ x\ cos\ y - cos \ x \ sin\ y}{sin \ x\ cos\ y + cos \ x \ sin \ y - sin \ x\ cos\ y + cos \ x \ sin\ y}\) 

=\(\cfrac{2a}{2b}\)

\(\cfrac{2\ sin\ x\ cos\ y }{2\ cos\ x\ sin\ y} \ = \cfrac{2a}{2b}\)

\(\cfrac{tan\ x}{tan\ y}= \cfrac{a}{b}\)

LATHAR

· commented

· 1 Months ago

option 3 also satisfies that
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