Reaching upon 9=1 while solving x for 3\tan(x−15^{\circ})=\tan(x+15^{\circ})Substituting y=x+45^{\circ}, we

Jessie Jenkins

Jessie Jenkins

Answered question

2022-01-24

Reaching upon 9=1 while solving x for 3tan(x15)=tan(x+15)
Substituting y=x+45, we get
3tan(y60)=tan(y30)
3tany31+3tany=tany131+13tany

3(tan2y3)=3tan21

9=1

The solution provided by the book x=nπ+π4 fits, so why did i get 9=1?

Answer & Explanation

Jaiden Conrad

Jaiden Conrad

Beginner2022-01-25Added 14 answers

You got
0tan2y+9=1
Id est, you got that tany does not exist.
Thus,
y=π2+πn
where nZ, or
x+45=π2+πn
which gives
x=πn+π4
liep3p

liep3p

Beginner2022-01-26Added 4 answers

Hint:
Set tany=1a
3(13a)a+3=3a3a+1
3a2=3(13a2)a=0

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