Question

Simplificando -se a expressão y = sen× /1+cos× + 1+cos×/ sen× obtém-se

Answer 1

$y=\dfrac{\sin{x}}{1+\cos{x}}+\dfrac{1+\cos{x}}{\sin{x}}\\\\\\y=\dfrac{(\sin{x})(\sin{x})+(1+\cos{x})(1+\cos{x})}{(1+\cos{x})(\sin{x})}\\\\\\y=\dfrac{(\sin^2x)+(1+2\cos{x}+\cos^2x)}{(1+\cos{x})(\sin{x})}\\\\\\y=\dfrac{\sin^2x+\cos^2x+1+2\cos{x}}{(1+\cos{x})(\sin{x})}\\\\\\y=\dfrac{1+1+2\cos{x}}{(1+\cos{x})(\sin{x})}\qquad\qquad\qquad\qquad\because\ \sin^2x+\cos^2x=1\\\\\\y=\dfrac{2+2\cos{x}}{(1+\cos{x})(\sin{x})}\\\\\\y=\dfrac{2(1+\cos{x})}{(1+\cos{x})(\sin{x})}\\\\\\\boxed{y=\dfrac{2}{\sin{x}}}$

Porém, utilizando uma conversão recursiva, ainda podemos prosseguir com a simplificação, embora, talvez irrelevante.

$y=2\cdot\dfrac{1}{\sin{x}}\\\\\\\boxed{y=2\,\text{cossec}\,x}$