Formulas for addition, double arc, half arc and transformation into product (chapters V and VI). Teste 75CESCEM-73 It is known that $\operatorname{tg} 75^\circ = 2 + \sqrt{3}$ and $\operatorname{tg} 60^\circ = \sqrt{3}$. The value of $\operatorname{tg} 15^\circ$ is: a) $\frac{1}{3}$b) $-\sqrt{3}$c) $\sqrt{3}$d) $2 + \sqrt{3}$e) $2 - \sqrt{3}$ $\operatorname{tg} 15^\circ = \operatorname{tg}(75^\circ - 60^\circ) = \frac{(2 + \sqrt{3}) - \sqrt{3}}{1 + (2 + \sqrt{3})\sqrt{3}} = \frac{2}{4 + 2\sqrt{3}} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$. Teste 76MACK-75 If $0 < a < \frac{\pi}{2}$ and $0 < b < \frac{\pi}{2}$, then: a) $\operatorname{sen}(a + b) < \operatorname{sen} a + \operatorname{sen} b$, whatever $a$ and $b$b) $\operatorname{sen}(a + b) > \operatorname{sen} a + \operatorname{sen} b$, whatever $a$ and $b$c) $\operatorname{sen}(a + b) > \operatorname{sen} a + \operatorname{sen} b$ only if $a > b$d) $\operatorname{sen}(a + b) < \operatorname{sen} a + \operatorname{sen} b$ only if $a < b$e) none of the above $\operatorname{sen}(a + b) = \operatorname{sen} a\cos b + \operatorname{sen} b\cos a$, and in the 1st quadrant (open) $0 < \cos a, \cos b < 1$; each installment is smaller than $\operatorname{sen} a$ and $\operatorname{sen} b$, respectively.