Question

Sabendo-se que sin⁡α=√5⁄5, sin⁡β=√10⁄10, 0<α<π/2 e 0<β<π/2, calcule o valor de tan⁡(α+β).

Answer 1

Antes de fazer tangente, precisamos descobrir o cosseno de cada um:

Sen²(α) + Cos²(α) = 1
5/25 + Cos²(α) = 1
Cos²(α) = 1 - 1/5
Cos²(α) = 4/5
Cos(α) = 2√5/5

Sen²(β) + Cos²(β) = 1
10/100 + Cos²(β) = 1
Cos²(β) = 1 - 1/10
Cos²(β) = 9/10
Cos(β) = 3√10/10

Feito isso, tangente é seno divido por cosseno:

$Tg(\alpha) = \frac{sen(\alpha)}{cos(\alpha)} = \frac{\sqrt{5}/5 }{2 \sqrt{5}/5 } \\ \\ Tg(\alpha) = \frac{\sqrt{5} } {5}. \frac{5}{2 \sqrt{5} } = \frac{1}{2}$

$Tg(\beta) = \frac{Sen(\beta)}{Cos(\beta)} = \frac{\sqrt{10}/10 }{3\sqrt{10}/10 } \\ \\ Tg(\beta) = \frac{\sqrt{10} }{10}. \frac{10}{3\sqrt{10} } = \frac{1}{3}$

Finalmente podemos calcular a soma das tangentes:

$Tg(\alpha + \beta) = \frac{Tg(\alpha) + Tg(\beta)}{1 - Tg(\alpha).Tg(\beta)} \\ \\ Tg(\alpha + \beta) = \frac{1/2 + 1/3}{1 - 1/2.1/3} = \frac{5/6}{5/6} = 1$

Answer 2

$\mathrm{\sin{\alpha}=\dfrac{\sqrt{5}}{5}\ \to\ \cos{\alpha}=\pm\sqrt{1-\sin^2{\alpha}}=\pm\sqrt{1-\bigg(\dfrac{\sqrt{5}}{5}\bigg)^2}=}\\\\ \mathrm{=\pm\sqrt{\dfrac{25}{25}-\dfrac{5}{25}}=\pm\sqrt{\dfrac{20}{25}}=\pm\dfrac{2\sqrt{5}}{5}\ \to\ \cos{\alpha}=\dfrac{2\sqrt{5}}{5}}\\\\ \mathrm{\tan{\alpha}=\dfrac{\sin{\alpha}}{\cos{\alpha}}=\dfrac{\frac{\sqrt{5}}{5}}{\frac{2\sqrt{5}}{5}}=\dfrac{\sqrt{5}}{5}.\dfrac{5}{2\sqrt{5}}=\dfrac{1}{2}\ \to\ \boxed{\mathrm{\tan{\alpha}=\dfrac{1}{2}}}}$

$\mathrm{\sin{\beta}=\dfrac{\sqrt{10}}{10}\ \to\ \cos{\beta}=\pm\sqrt{1-\sin^2{\beta}}=\pm\sqrt{1-\bigg(\dfrac{\sqrt{10}}{10}\bigg)^2}=}\\\\ \mathrm{=\pm\sqrt{\dfrac{100}{100}-\dfrac{10}{100}}=\pm\sqrt{\dfrac{90}{100}}=\pm\dfrac{3\sqrt{10}}{10}\ \to\ \cos{\beta}=\dfrac{3\sqrt{10}}{10}}\\\\ \mathrm{\tan{\beta}=\dfrac{\sin{\beta}}{\cos{\beta}}=\dfrac{\frac{\sqrt{10}}{10}}{\frac{3\sqrt{10}}{10}}=\dfrac{\sqrt{10}}{10}.\dfrac{10}{3\sqrt{10}}=\dfrac{1}{3}\ \to\ \boxed{\mathrm{\tan{\beta}=\dfrac{1}{3}}}}$

$\mathrm{\tan{(\alpha+\beta)}=\dfrac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}=\dfrac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}=\dfrac{\frac{3+2}{6}}{\frac{6}{6}-\frac{1}{6}}=\dfrac{\frac{5}{6}}{\frac{5}{6}}=1}\\\\\\ \boxed{\boxed{\mathbf{\tan{(\alpha+\beta)=1}}}}$