Question

Someone help. me solving this​

Answer 1

Hope It Helps
The answer would be This

Answer 2

Step-by-step explanation:

tany = 3tanx

tanx = tan y /3 ->1

tan(x+y) = tanx + tany

1- tanx tany

from 1

tan(x+y = tany + tany

3 ( from 1)

1- tany tany

3

on taking LCM

tan(x+y) = tany + 3tany

3 - tan2y

= 4tany

3- tan2y

= 4siny

cosy

3 - sin 2y

cos2y

On taking LCM in the denominator and canceling one cosy in the denominator's denominator

tan ( x + y) = 4siny / cosy

3cos2y - Sin2y

cos2y

=>

4sinycosy

2cos2y + cos2y - sin2y

=>

2(2sinycosy)

cos2y +1 +co2y ( cos2y + 1 = 2cos2y and cos2y - sin2y = cos2y)

=>

2sin2y

1 + 2cosy