Instrucciones: Responde las siguientes preguntas.

• Utiliza la fórmula de la tangente de la diferencia de dos ángulos para calcular a $\tan {10}^{\circ }$. Considera que $\tan {50}^{\circ }\approx 1.1918$ y $\tan {40}^{\circ }\approx 0.8391$.
  • $$\tan \left({10}^{\circ }\right)\approx \frac{1+\left(1.1918\right)\left(0.8391\right)}{1.1918-0.8391}\approx 5.6707$$
  • $$\tan \left({10}^{\circ }\right)\approx \frac{1.1918-0.8391}{1+\left(1.1918\right)\left(0.8391\right)}\approx 0.1763$$
  • $$\tan \left({10}^{\circ }\right)\approx \frac{1.1918+0.8391}{2\left(1.1918\right)\left(0.8391\right)}\approx 1.0154$$
  • $$\tan \left({10}^{\circ }\right)\approx \frac{2\left(1.1918\right)\left(0.8391\right)}{1.1918+0.8391}\approx 0.9848$$

• Utiliza la fórmula de la tangente de la suma de dos ángulos para calcular a $\tan {22}^{\circ }$. Considera que $\tan {10}^{\circ }\approx 0.1763$ y $\tan {12}^{\circ }\approx 0.2126$.
  • $$\tan \left({22}^{\circ }\right)\approx \frac{0.1763+0.2126}{1-\left(0.1763\right)\left(0.2126\right)}\approx 0.4040$$
  • $$\tan \left({22}^{\circ }\right)\approx \frac{2\left(0.2126\right)\left(0.1763\right)}{0.2126+0.1763}\approx 0.1928$$
  • $$\tan \left({22}^{\circ }\right)\approx \frac{1-\left(0.1763\right)\left(0.2126\right)}{0.1763+0.2126}\approx 2.4750$$
  • $$\tan \left({22}^{\circ }\right)\approx \frac{0.2126+0.1763}{2\left(0.2126\right)\left(0.1763\right)}\approx 5.1879$$

• Identifica la identidad trigonométrica correcta.
  • $$\tan (\alpha +\beta )=\frac{1-\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }$$
  • $$\tan (\alpha +\beta )=\frac{2\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }$$
  • $$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$$
  • $$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{2\tan \alpha \tan \beta }$$

• Identifica la identidad trigonométrica correcta.
  • $$\tan (\alpha -\beta )=\frac{2\tan \alpha \tan \beta }{\tan \alpha +\tan \beta }$$
  • $$\tan (\alpha +\beta )=\frac{\tan \alpha -\tan \beta }{2\tan \alpha \tan \beta }$$
  • $$\tan (\alpha -\beta )=\frac{1+\tan \alpha \tan \beta }{\tan \alpha -\tan \beta }$$
  • $$\tan (\alpha -\beta )=\frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }$$

• Utiliza la fórmula de la tangente de la suma de dos ángulos para calcular a $\tan {80}^{\circ }$. Considera que $\tan {25}^{\circ }\approx 0.4663$ y $\tan {55}^{\circ }\approx 1.4281$.
  • $$\tan \left({80}^{\circ }\right)\approx \frac{1-\left(0.4663\right)\left(1.4281\right)}{0.4663+1.4281}\approx 0.1763$$
  • $$\tan \left({80}^{\circ }\right)\approx \frac{2\left(0.4663\right)\left(1.4281\right)}{1.4281+0.4663}\approx 0.7030$$
  • $$\tan \left({80}^{\circ }\right)\approx \frac{0.4663+1.4281}{1-\left(0.4663\right)\left(1.4281\right)}\approx 5.6705$$
  • $$\tan \left({80}^{\circ }\right)\approx \frac{1.4281+0.4663}{2\left(0.4663\right)\left(1.4281\right)}\approx 1.4224$$