Evaluate tan (110°)-tan(50°)/1+tan110°×tan50°
First, we need to calculate the values of $\tan(110°)$ and $\tan(50°)$.
$\tan(110°) = \frac{\sin(110°)}{\cos(110°)}$
By using the identity $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$:
$\sin(110°) = \sin(90° + 20°) = \sin(90°) \cos(20°) + \cos(90°)\sin(20°) = \cos(20°)$
$\cos(110°) = \cos(90° + 20°) = \cos(90°) \cos(20°) - \sin(90°)\sin(20°) = -\sin(20°)$
Therefore, $\tan(110°) = \frac{\cos(20°)}{-\sin(20°)}=-\cot(20°)$
Next, $\tan(50°) = \frac{\sin(50°)}{\cos(50°)}$
By using the identity $\sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b)$:
$\sin(50°) = \sin(30° + 20°) = \sin(30°) \cos(20°) + \cos(30°)\sin(20°) = \frac{1}{2} \cos(20°) + \frac{\sqrt{3}}{2} \sin(20°)$
$\cos(50°) = \cos(30° + 20°) = \cos(30°) \cos(20°) - \sin(30°)\sin(20°) = \frac{\sqrt{3}}{2} \cos(20°) - \frac{1}{2} \sin(20°)$
Therefore, $\tan(50°) = \frac{\frac{1}{2} \cos(20°) + \frac{\sqrt{3}}{2} \sin(20°)}{\frac{\sqrt{3}}{2} \cos(20°) - \frac{1}{2} \sin(20°)}$
Now we can substitute these values back into the original expression:
$\frac{-\cot(20°) - \tan(50°)}{1 + \cot(20°) \tan(50°)}$
This expression can be further simplified using trigonometric identities and algebraic manipulation.