Question

Dado α = 30 º e β = 45º , encontre Tg (α+ β) : 

a) 9

b) 7

c) 5

d) 1

e) 4

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Answer 1

$\tan(\alpha+\beta) = \dfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}\\~\\\tan(\alpha+\beta) =\dfrac{\sin{\alpha}\cos{\beta}+\sin{\beta}\cos{\alpha}}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}\\~$

$\tan(30+45) =\dfrac{\sin{30}\cos{45}+\sin{45}\cos{30}}{\cos30\cos45-\sin30\sin45}\\~$

$\beta = 45 \implies \cos\beta = \sin\beta$

$\tan(30+45) =\dfrac{\sin{30}\cos{45}+\sin{45}\cos{30}}{\cos30\cos45-\sin30\sin45}\\~\\\tan(30+45) =\dfrac{\sin{30}+\cos{30}}{\cos30-\sin30}\\~\\\tan(30+45) =\dfrac{1+\sqrt{3}}{\sqrt{3}-1}\\~\\\tan(30+45) =2+\sqrt{3}$

aproximadamente e) 4