Question

o produto (sec x - cos x) (tg x + cotg x) é identico a:

Answer 1

$(sec x-\cos x)(\tan x+cotg x)=\\\\(\frac{1}{\cos x}-\cos x)(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})=\\\\\\(\frac{1-\cos^2 x}{\cos x})(\frac{\sin^2 x+\cos^2 x}{\sin x\cdot\cos x})=$

 Lembre-se que: $\sin^2x+\cos^2x=1$.

$(\frac{1-\cos^2x}{\cos x})(\frac{\sin^2x+\cos^2x}{\sin x\cdot\cos x})=\\\\\\(\frac{\sin^2x}{\cos x})(\frac{1}{\sin x\cdot\cos x})=\\\\\\\frac{\sin^2x}{\sin x\cdot\cos^2 x}=\\\\\\\frac{\sin x}{\cos^2x}=\\\\\\\frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}=\\\\\ \boxed{\tan x\cdot sec\;x}$

 Parece-me que deixaste de postar as alternativas!