Question

sabendo que a tg a=$\frac{1}{2}$ e sen b=$\frac{2}{5}$, com a e b pertencentes ao primeiro quadrante, calcule tg(a+b)

Answer 1

$sen^2(b)+cos^2(b)=1$

$( \frac{2}{5} )^2+cos^2(x)=1,\ portanto\ cos(x)= \sqrt{1- \frac{4}{25} }=\sqrt{\frac{21}{25} }= \frac{ \sqrt{21} }{5}$

$t(b)= \frac{sen(b)}{cos(b)}= \frac{ \frac{2}{5} }{ \frac{ \sqrt{21} }{5} } = \frac{ 2\sqrt{21} }{21}$

$tg(a+b)= \frac{ \frac{1}{2} + \frac{2 \sqrt{21} }{21} }{1-\frac{1}{2}*\frac{2 \sqrt{21} }{21}} =\frac{ \frac{21+4 \sqrt{21} }{42} }{\frac{21-\sqrt{21} }{21}}= \frac{21+4 \sqrt{21} }{2(21-\sqrt{21})}$


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