De acordo com o assunto Matemática: Razões e Funções Trigonométricas. Ciclo Trigonométrico., responda: Simplifique a expressão \( \dfrac{\sin x+\cot x}{\csc x\cdot\tan x}\) em term

\( \dfrac{\cos x\big(\sin^2 x+\cos x\big)}{\sin x}\)

Explicação:
Usar definições: \(\cot x=\dfrac{\cos x}{\sin x}\), \(\tan x=\dfrac{\sin x}{\cos x}\), \(\csc x=\dfrac{1}{\sin x}\).
Denominador: \(\csc x\cdot\tan x=\dfrac{1}{\sin x}\cdot\dfrac{\sin x}{\cos x}=\dfrac{1}{\cos x}\).
Numerador: \(\sin x+\cot x=\sin x+\dfrac{\cos x}{\sin x}=\dfrac{\sin^2 x+\cos x}{\sin x}\).
Portanto
\(\dfrac{\sin x+\cot x}{\csc x\cdot\tan x}=\dfrac{\dfrac{\sin^2 x+\cos x}{\sin x}}{1/\cos x}=\dfrac{\cos x(\sin^2 x+\cos x)}{\sin x}.\)

Exemplo:
Para \(x=45^\circ\) (\(\pi/4\)): \(\sin x=\cos x=\dfrac{\sqrt{2}}{2}\).
\(\sin^2 x=\tfrac12\),\; \(\sin^2 x+\cos x=\tfrac12+\tfrac{\sqrt{2}}{2}=\tfrac{1+\sqrt{2}}{2}\).
Aplicando a fórmula:
\(\dfrac{\cos x(\sin^2 x+\cos x)}{\sin x}=\dfrac{\tfrac{\sqrt{2}}{2}\cdot\tfrac{1+\sqrt{2}}{2}}{\tfrac{\sqrt{2}}{2}}=\dfrac{1+\sqrt{2}}{2}.\)