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Let P(x, y) be the terminal point on the unit circle determined by t. Then sint=,cost=, and tant=.
Reaching upon 9=1 while solving x for 3tan(x−15∘)=tan(x+15∘) Substituting y=x+45∘, we get 3tan(y−60∘)=tan(y−30∘) 3tany−31+3tany=tany−131+13⋅tany
3(tan2y−3)=3tan2−1
9=1
The solution provided by the book x=nπ+π4 fits, so why did i get 9=1?
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