Question

sabendo que cos(x)=3/5 qual o valor de 9sec^2(x)+4cotg(x)?

Answer 1

$\boxed{\begin{array}{l}\sf vou~assumir~que~x\in~1^o~quadrante\\\sf cos(x)=\dfrac{3}{5}\longrightarrow sec(x)=\dfrac{5}{3}\\\sf tg^2(x)=sec^2(x)-1\\\sf tg^2(x) =\bigg(\dfrac{5}{3}\bigg)^2-1\\\sf tg^2(x)=\dfrac{25}{9}-1\\\sf tg^2(x)=\dfrac{16}{9}\\\sf tg(x)=\sqrt{\dfrac{16}{9}} \\\sf tg(x)=\dfrac{4}{3}\\\sf cotg(x)=\dfrac{3}{4}\\\sf 9sec^2(x)+4cotg(x)=\diagup\!\!\!9\cdot\dfrac{25}{\diagup\!\!\!9}+\diagup\!\!\!\!4\cdot\dfrac{3}{\diagup\!\!\!\!4}\\\sf 9sec^2(x)+4cotg(x)=25+3\\\sf 9sec^2(x)+4cotg(x)=28\end{array}}$