Question

demonstre a identidade tgx + cotgx=tg x . cossec2x

Answer 1

É cossec² x  ?   Se for, segue a demonstração:

tg x + cotg x = sen x / cos x  +  cos x / sen x = (sen² x + cos² x) / sen x . cos x = 1 / sen x . cos x   (*)

tg x . cossec² x = sen x / cos x  .  1 / sen² x = 1 / sen x . cos x (**)

Como (*) = (**), está demonstrada a identidade.

Answer 2

$\large\begin{array}{l} \textsf{Mostrar que }\\\\ \mathsf{tg\,x+cotg\,x=tg\,x\cdot cossec^2\,x} \end{array}$


$\large\begin{array}{l} \textsf{A ideia envolvida \'e partir do lado esquerdo da igualdade}\\\textsf{e chegar ao lado direito. Observe:}\\\\ \mathsf{tg\,x+cotg\,x}\\\\ =\mathsf{tg\,x+\dfrac{1}{tg\,x}}\\\\ =\mathsf{tg\,x+\dfrac{tg\,x}{tg^2\,x}}\\\\ =\mathsf{tg\,x\cdot \left(1+\dfrac{1}{tg^2\,x}\right)}\\\\ =\mathsf{tg\,x\cdot \left(1+cotg^2\,x\right)}\\\\ \end{array}$

$\large\begin{array}{l} =\mathsf{tg\,x\cdot \left(1+\dfrac{cos^2\,x}{sen^2\,x}\right)}\\\\ =\mathsf{tg\,x\cdot \left(\dfrac{sen^2\,x}{sen^2\,x}+\dfrac{cos^2\,x}{sen^2\,x}\right)}\\\\ =\mathsf{tg\,x\cdot \left(\dfrac{sen^2\,x+cos^2\,x}{sen^2\,x}\right)}\\\\ =\mathsf{tg\,x\cdot \left(\dfrac{1}{sen^2\,x}\right)}\\\\ =\mathsf{tg\,x\cdot cossec^2\,x\qquad\checkmark} \end{array}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$