Question

Demonstre a relação verdade da identidade trigonométrica tg(45º+x) = cos(x) + sen(x) / cos(x) - sen (x)

Answer 1

$\boxed{\boxed{tg(a\pm b)=\dfrac{t(a)\pm\tg(b)}{1\mp tg(a)\cdot tg(b)}}}$

Então:

$tg(45\º+x)=\dfrac{tg(45\º)+tg(x)}{1-tg(45\º)\cdot tg(x)}\\\\\\tg(45\º+x)=\dfrac{1+tg(x)}{1-1\cdot tg(x)}\\\\\\tg(45\º+x)=\dfrac{1+tg(x)}{1-tg(x)}\\\\\\tg(45\º+x)=\dfrac{1+(\frac{sen(x)}{cos(x)})}{1-(\frac{sen(x)}{cos(x)})}\\\\\\tg(45\º+x)=\dfrac{(\frac{cos(x)+sen(x)}{cos(x)})}{(\frac{cos(x)-sen(x)}{cos(x)})}\\\\\\tg(45\º+x)=\dfrac{cos(x)-sen(x)}{cos(x)}\cdot\dfrac{cos(x)}{cos(x)-sen(x)}\\\\\\\boxed{\boxed{tg(45\º+x)=\dfrac{cos(x)+sen(x)}{cos(x)-sen(x)}~~c.q.d}}$