Use the identity for tan(A+B) or tan(A-B) to solve each problem {1-6}
1) tan75° = tan(30° + 45°)
2) tan15° = tan(45° + 30°)
3) tan105° = tan(60° + 45°)
4) tan195° = tan(135° + 60°)
5) tan165° = tan(135° + 30°)
6) tan(-7π/12)=
+36916 The Identity is:
Tan (A+B) = (Tan A + Tan B) / (1-TanA TanB)
I'll do the first one....you can do the rest..I'll use exact values rather than calculator values
Tan(30+45) = (tan30 + Tan 45) / (1-(tanA tanB )
= (1/2/(sqrt3/2) +1)/ (1 -(1/2(sqrt3/2)(1))
= (1 + 1/sqrt3) / )(1-(sqrt3))
((sqrt3 +1)sqrt3) / (sqrt3-1)/sqrt3
(sqrt3+1) / ( sqrt3-1) ~ = 3.73
+128460 Thanks, EP
The last one might be a little difficult
-7pi/ 12 x 180 / pi = (-7/12)* 180 = -105° = 360 - 105 = 255°
And 255° = 210° + 45°
So
tan ( 255°) = tan ( 210° + 45°) =
tan (210) + tan (45) 1/√3 + 1 [ √3 + 1 ] / √3
_________________ = _______________ = ______________ =
1 - tan (210)tan(45) 1 - (1/√3)(1) 1 - 1/√3
[ √3 + 1 ] / √3 [ √3 + 1 ] 4 + 2√3
________________ = ____________ = ________ = 2 + √3
[ √3 - 1 ] / √3 [ √3 - 1 ] 3 - 1
