# How do you solve \frac { \tan \theta - \tan 27^ { \circ } } { 1+ \tan \theta \tan 27^ { \circ } } = 1?

Nov 13, 2017

$\setminus \theta = {72}^{\circ} + {180}^{\circ} \mathmr{and} \setminus \theta = \frac{2 \pi}{5} + n \setminus \pi$ where $n \in \mathbb{Z}$

#### Explanation:

$\frac{\tan \setminus \theta - \tan {27}^{\circ}}{1 + \tan \setminus \theta \cdot \tan {27}^{\circ}} = 1$

Using the trig identity: $\tan \left(A - B\right) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, we can get:
$\frac{\tan \setminus \theta - \tan {27}^{\circ}}{1 + \tan \setminus \theta \cdot \tan {27}^{\circ}} = \tan \left(\setminus \theta - {27}^{\circ}\right) = 1$

$\arctan \left(\tan \left(\setminus \theta - {27}^{\circ}\right)\right) = \setminus \theta - {27}^{\setminus} \circ$
$\arctan \left(1\right) = {45}^{\circ}$

$\setminus \theta - {27}^{\circ} = {45}^{\circ}$

$\setminus \theta = {45}^{\circ} + {27}^{\circ} = {72}^{\circ}$

$\setminus \theta = {72}^{\circ} + n {180}^{\circ} \mathmr{and} \setminus \theta = \frac{2 \pi}{5} + n \pi$ where $n \in \mathbb{Z}$

Nov 13, 2017

$\setminus \theta = {72}^{\setminus} \circ$

#### Explanation:

$\frac{\tan \setminus \theta - \tan \left(27\right)}{1 + \tan \setminus \theta \tan \left(27\right)} = 1$

$\frac{\left(\tan \setminus \theta - \tan \left(27\right)\right) \cancel{\left(1 + \tan \setminus \theta \tan \left(27\right)\right)}}{\cancel{1 + \tan \setminus \theta \tan \left(27\right)}} = 1 \cdot \left(1 + \tan \setminus \theta \tan \left(27\right)\right)$

$\tan \setminus \theta - \tan \left(27\right) = 1 + \tan \setminus \theta \tan \left(27\right)$

$\tan \setminus \theta - \tan \setminus \theta \tan \left(27\right) = 1 + \tan \left(27\right)$

$\tan \setminus \theta \left(1 - \tan \left(27\right)\right) = \left(1 + \tan \left(27\right)\right)$

$\tan \setminus \theta = \frac{1 + \tan \left(27\right)}{1 - \tan \left(27\right)}$

$\setminus \theta = \arctan \left(\frac{1 + \tan \left(27\right)}{1 - \tan \left(27\right)}\right) \approx \arctan \left(3.08\right) \approx {72}^{\setminus} \circ$

Nov 14, 2017

$\theta = {72}^{\circ}$

#### Explanation:

$\frac{\tan \setminus \theta - \tan {27}^{\circ}}{1 + \tan \setminus \theta \cdot \tan {27}^{\circ}} = 1$

$\tan \setminus \theta - \tan {27}^{\circ} = 1 + \tan \setminus \theta \cdot \tan {27}^{\circ}$

$\tan \setminus \theta - \tan \setminus \theta \cdot \tan {27}^{\circ} = 1 + \tan {27}^{\circ}$

$\tan \setminus \theta \cdot \left(1 - \tan {27}^{\circ}\right) = 1 + \tan {27}^{\circ}$

$\tan \setminus \theta = \frac{1 + \tan {27}^{\circ}}{1 - \tan {27}^{\circ}}$

=$\frac{\cos {27}^{\circ} + \sin {27}^{\circ}}{\cos {27}^{\circ} - \sin {27}^{\circ}}$

=$\frac{\sin {63}^{\circ} + \sin {27}^{\circ}}{\cos {27}^{\circ} - \cos {63}^{\circ}}$

=$\frac{2 \sin {45}^{\circ} \cdot \cos {18}^{\circ}}{- 2 \sin {45}^{\circ} \cdot \sin {\left(- 18\right)}^{\circ}}$

=$\cos {18}^{\circ} / - \left(- \sin {18}^{\circ}\right)$

=$\cot {18}^{\circ}$

=$\tan {72}^{\circ}$

Hence $\theta = {72}^{\circ}$