Question

Simplifique tgx+cotgx/secx.cotgx

Answer 1

$\large\begin{array}{l} \mathsf{\dfrac{tg\,x+cotg\,x}{sec\,x\cdot cotg\,x}}\\\\ =\mathsf{\dfrac{\frac{sen\,x}{cos\,x}+\frac{cos\,x}{sen\,x}}{\frac{1}{\;\diagup\!\!\!\!\!\!\! cos\,x}\cdot \frac{\;\diagup\!\!\!\!\!\!\! cos\,x}{sen\,x}}}\\\\ =\mathsf{\dfrac{\frac{sen\,x}{cos\,x}+\frac{cos\,x}{sen\,x}}{\frac{1}{sen\,x}}}\\\\=\mathsf{\dfrac{\frac{sen^2\,x}{sen\,x\cdot cos\,x}+\frac{cos^2\,x}{sen\,x\cdot cos\,x}}{\frac{1}{sen\,x}}}\\\\ =\mathsf{\dfrac{sen^2\,x+cos^2\,x}{\;\diagup\!\!\!\!\!\!\! sen\,x\cdot cos\,x}\cdot \;\diagup\!\!\!\!\!\!\! sen\,x}\qquad\quad\textsf{(mas }\mathsf{sen^2\,x+cos^2\,x=1}\textsf{)} \end{array}$

$\large\begin{array}{l} =\mathsf{\dfrac{1}{cos\,x}}\\\\ =\mathsf{sec\,x\qquad\checkmark} \end{array}$


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$\large\textsf{Bons estudos! :-)}$