Question

Exercițiul 17 ??????

Answer 1

Răspuns:

tgx=2√6

Explicație pas cu pas:

Aplici     formula    sin²x+c0s²x=1

sin²x+(1/5)²=1

sin²x+1/25=1

sin²x=24/25

sinx=√24/25=√4*6/25=2√6/5

tgx=sinx/cosx=[2√6/5]/(1/5)=2√6

Answer 2

$\it x\in\Big[0,\ \dfrac{\pi}{2}\Big]\ \Rightarrow\ tgx\geq0\\ \\ \\ tgx=\dfrac{sinx}{cosx} \Rightarrow tg^2x=\dfrac{sin^2x}{cos^2x}=\dfrac{1-cos^2x}{cos^2x}=\dfrac{1}{cos^2x}-1=\dfrac{1}{\dfrac{1}{25}}-1=\\ \\ \\ =25-1=24 \Rightarrow tgx=\sqrt{24}=\sqrt{4\cdot6}=2\sqrt6$