Question

sendo sen x + cos x = a,calule tg x + cotg x

Answer 1

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$\large\begin{array}{l} \textsf{Se}\\\\ \mathsf{sen\,x+cos\,x=a}\\\\ \textsf{onde }\mathsf{sen\,x\ne 0~~e~~cos\,x\ne 0,}\textsf{ ent\~ao} \end{array}$


$\large\begin{array}{l} \mathsf{tg\,x+cotg\,x}\\\\ =\mathsf{\dfrac{sen\,x}{cos\,x}+\dfrac{cos\,x}{sen\,x}}\\\\ =\mathsf{\dfrac{sen^2\,x}{sen\,x\,cos\,x}+\dfrac{cos^2\,x}{sen\,x\,cos\,x}}\\\\ =\mathsf{\dfrac{sen^2\,x+cos^2\,x}{sen\,x\,cos\,x}} \end{array}$

$\large\begin{array}{l} =\mathsf{\dfrac{1}{sen\,x\,cos\,x}}\\\\ =\mathsf{\dfrac{2}{2\,sen\,x\,cos\,x}}\\\\ =\mathsf{\dfrac{2}{2\,sen\,x\,cos\,x+1-1}}\qquad\textsf{(mas }\mathsf{1=sen^2\,x+cos^2\,x}\textsf{)}\\\\ =\mathsf{\dfrac{2}{2\,sen\,x\,cos\,x+sen^2x+cos^2x-1}} \end{array}$

$\large\begin{array}{l} =\mathsf{\dfrac{2}{(sen^2\,x+2\,sen\,x\,cos\,x+cos^2x)-1}}\\\\ =\mathsf{\dfrac{2}{(sen\,x+cos\,x)^2-1}}\\\\ =\mathsf{\dfrac{2}{a^2-1}\qquad\quad\checkmark} \\\\\\ \therefore~~\boxed{\begin{array}{c}\mathsf{tg\,x+cotg\,x=\dfrac{2}{a^2-1}}\end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}$


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


Tags:  identidade transformação trigonométrica fatoração produtos notáveis quadrado da soma seno cosseno tangente cotangente sen cos tg tan cotg cot trigonometria