(1) Determina el valor de cada función utilizando las identidades para suma \( y \) diferencia de ángulos. \( \begin{array}{ll}\text { a. } \cos 165^{\circ} & \text { h. } \tan 105^{\circ} \\ \text { b. } \operatorname{sen} 15^{\circ} & \text { i. } \cos 195^{\circ} \\ \text { c. } \operatorname{sen} 75^{\circ} & \text { j. } \operatorname{sen} 285^{\circ} \\ \text { d. } \cos 345^{\circ} & \text { k. } \tan \left(-120^{\circ}\right) \\ \text { c. } \operatorname{sen}\left(-\frac{7}{12} \pi\right) & \text { l. } \cos \frac{23}{12} \pi \\ \text { f. } \tan \frac{13}{12} \pi & \text { mi. } \operatorname{sen} \frac{19}{12} \pi \\ \text { g. } \csc \frac{5}{12} \pi & \text { n. } \tan \frac{17}{12} \pi\end{array} \)

Claro, aquí tienes las soluciones para cada una de las funciones utilizando las identidades para suma y diferencia de ángulos: a. \(\cos 165^{\circ}\) \[ \cos 165^{\circ} = \cos (180^{\circ} - 15^{\circ}) = -\cos 15^{\circ} \] Usando la identidad \(\cos (90^{\circ} - \theta) = \sin \theta\): \[ -\cos 15^{\circ} = -\sin 75^{\circ} \] Y usando la identidad \(\sin (90^{\circ} - \theta) = \cos \theta\): \[ -\sin 75^{\circ} = -\cos 15^{\circ} \] Finalmente, usando el valor de \(\cos 15^{\circ}\): \[ -\cos 15^{\circ} = -\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) \] b. \(\sin 15^{\circ}\) \[ \sin 15^{\circ} = \sin (45^{\circ} - 30^{\circ}) \] Usando la identidad \(\sin (A - B) = \sin A \cos B - \cos A \sin B\): \[ \sin 15^{\circ} = \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \] \[ \sin 15^{\circ} = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \] \[ \sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4} \] c. \(\sin 75^{\circ}\) \[ \sin 75^{\circ} = \sin (90^{\circ} - 15^{\circ}) = \cos 15^{\circ} \] Usando el valor de \(\cos 15^{\circ}\): \[ \cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4} \] d. \(\cos 345^{\circ}\) \[ \cos 345^{\circ} = \cos (360^{\circ} - 15^{\circ}) = \cos 15^{\circ} \] Usando el valor de \(\cos 15^{\circ}\): \[ \cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4} \] e. \(\tan \left(-120^{\circ}\right)\) \[ \tan \left(-120^{\circ}\right) = -\tan 120^{\circ} \] Usando la identidad \(\tan (-\theta) = -\tan \theta\): \[ -\tan 120^{\circ} = -\tan (180^{\circ} - 60^{\circ}) = -\tan 60^{\circ} \] \[ -\tan 60^{\circ} = -\sqrt{3} \] f. \(\tan \frac{13}{12} \pi\) \[ \tan \frac{13}{12} \pi = \tan \left(\pi + \frac{1}{12} \pi\right) = \tan \frac{1}{12} \pi \] Usando la identidad \(\tan (\pi + \theta) = \tan \theta\): \[ \tan \frac{1}{12} \pi = \tan \left(\frac{\pi}{3} - \frac{\pi}{4}\right) \] Usando la identidad \(\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\): \[ \tan \left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\tan \frac{\pi}{3} - \tan \frac{\pi}{4}}{1 + \tan \frac{\pi}{3} \tan \frac{\pi}{4}} \] \[ \tan \left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\sqrt{3} - 1}{1 + \sqrt{3} \cdot 1} \] \[ \tan \left(\frac{\pi}{3} - \frac{\pi}{4}\right) = \frac{\sqrt{