Instrucciones: Responde las siguientes preguntas.

• 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 \(\tan 12^{\circ} \approx 0.2126\).
  • $$\tan \left ( 22^{\circ} \right ) \approx \frac{0.2126+0.1763}{2\left ( 0.2126 \right )\left ( 0.1763 \right )} \approx 5.1879$$
  • $$\tan \left ( 22^{\circ} \right ) \approx \frac{2\left ( 0.2126 \right )\left ( 0.1763 \right )}{0.2126+0.1763} \approx 0.1928$$
  • $$\tan \left ( 22^{\circ} \right ) \approx \frac{1- \left ( 0.1763 \right )\left ( 0.2126 \right )}{0.1763+0.2126} \approx 2.4750$$
  • $$\tan \left ( 22^{\circ} \right ) \approx \frac{0.1763+0.2126}{1- \left ( 0.1763 \right )\left ( 0.2126 \right )} \approx 0.4040$$

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

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

• 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 \(\tan 55^{\circ} \approx 1.4281\).
  • $$\tan \left ( 80^{\circ} \right ) \approx \frac{1.4281+0.4663}{2\left ( 0.4663 \right )\left ( 1.4281 \right )} \approx 1.4224$$
  • $$\tan \left ( 80^{\circ} \right ) \approx \frac{0.4663+1.4281}{1- \left ( 0.4663 \right )\left ( 1.4281 \right )} \approx 5.6705$$
  • $$\tan \left ( 80^{\circ} \right ) \approx \frac{2\left ( 0.4663 \right )\left ( 1.4281 \right )}{1.4281+0.4663} \approx 0.7030$$
  • $$\tan \left ( 80^{\circ} \right ) \approx \frac{1- \left ( 0.4663 \right )\left ( 1.4281 \right )}{0.4663+1.4281} \approx 0.1763$$

• 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 \(\tan 40^{\circ} \approx 0.8391\).
  • $$\tan \left ( 10^{\circ} \right ) \approx \frac{1+ \left ( 1.1918 \right )\left ( 0.8391 \right )}{1.1918-0.8391} \approx 5.6707$$
  • $$\tan \left ( 10^{\circ} \right ) \approx \frac{1.1918+0.8391}{2\left ( 1.1918 \right )\left ( 0.8391 \right )} \approx 1.0154$$
  • $$\tan \left ( 10^{\circ} \right ) \approx \frac{2\left ( 1.1918 \right )\left ( 0.8391 \right )}{1.1918+0.8391} \approx 0.9848$$
  • $$\tan \left ( 10^{\circ} \right ) \approx \frac{1.1918-0.8391}{1+ \left ( 1.1918 \right )\left ( 0.8391 \right )} \approx 0.1763$$