Question

Daca sin x= 3/5 si 0< x < $\pi /2$ , atunci $\frac{ctgx+tgx}{ctgx-tgx}$ =?

Answer 1

[tex]\text{Stim ca:}\\ tg\ x=\dfrac{\sin x}{\cos x}=\dfrac{\sin x}{\sqrt{1-\sin^2 x}}\\ \text{Deci vom avea ca:}\\ tg\ x=\dfrac{\frac{3}{5}}{\sqrt{1-\frac{9}{25}}}=\dfrac{\frac{3}{5}}{\sqrt{\frac{16}{25}}}=\dfrac{3}{5} \cdot \dfrac{ 5}{4}=\boxed{\dfrac{3}{4}}\\ \text{Din relatia:}tg\ x\cdot ctg\ x=1\Rightarrow ctg\ x=\dfrac{1}{tg\ x}\\ \text{Deci }ctg\ x=\dfrac{4}{3}\\ [/tex]
$\text{Asadar:}\dfrac{ctg\ x+tg\ x}{ctg\ x-tg\ x}=\dfrac{\frac{3}{4}+\frac{4}{3}}{\frac{4}{3}- \frac{3}{4}}=\dfrac{3\cdot 3+4\cdot 4}{12}\cdot \dfrac{12}{4\cdot 4-3\cdot 3}=\boxed{\dfrac{25}{7}}$