Escreva \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x} em termos de \sin x.

Passo 1

O enunciado pede para reescrever
\frac{1}{1-\cos x}\;-\;\frac{\cos x}{1+\cos x}
usando apenas \sin x. O conteúdo principal é manipular identidades trigonométricas, em especial as de ângulo metade e a relação do seno de ângulo duplo.

Passo 2

Usamos as identidades de meia-ângulo:
1 - \cos x = 2\,\sin^2\!\left(\frac{x}{2}\right), \quad 1 + \cos x = 2\,\cos^2\!\left(\frac{x}{2}\right).

Passo 3

Reescrevemos cada fração substituindo os denominadores:
\frac{1}{1-\cos x} = \frac{1}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)}, \qquad \frac{\cos x}{1+\cos x} = \frac{\cos x}{2\,\cos^2\!\bigl(\tfrac{x}{2}\bigr)} = \frac{1}{2}\,\frac{\cos x}{\cos^2\!\bigl(\tfrac{x}{2}\bigr)}.

Passo 4

Observando que \cos x = 2\cos^2\!\bigl(\tfrac{x}{2}\bigr)-1, temos
\frac{\cos x}{1+\cos x} = \frac{2\cos^2\!\bigl(\tfrac{x}{2}\bigr)-1}{2\,\cos^2\!\bigl(\tfrac{x}{2}\bigr)} = 1 \;-\;\frac{1}{2\,\cos^2\!\bigl(\tfrac{x}{2}\bigr)}.
Logo a expressão original vira
\frac{1}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)} \;-\;\Bigl(1 - \frac{1}{2\,\cos^2\!\bigl(\tfrac{x}{2}\bigr)}\Bigr) = \frac{1}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)} -1 + \frac{1}{2\,\cos^2\!\bigl(\tfrac{x}{2}\bigr)}.

Passo 5

Somamos as frações em \sin^2(\tfrac{x}{2}) e \cos^2(\tfrac{x}{2}):
\frac{1}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)} + \frac{1}{2\,\cos^2\!\bigl(\tfrac{x}{2}\bigr)} = \frac{\sin^2\!\bigl(\tfrac{x}{2}\bigr)+\cos^2\!\bigl(\tfrac{x}{2}\bigr)}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)\cos^2\!\bigl(\tfrac{x}{2}\bigr)} = \frac{1}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)\cos^2\!\bigl(\tfrac{x}{2}\bigr)}.
Portanto a expressão fica
\frac{1}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)\cos^2\!\bigl(\tfrac{x}{2}\bigr)} -1.

Passo 6

Usamos a identidade do seno de ângulo duplo:
\sin x = 2\,\sin\!\bigl(\tfrac{x}{2}\bigr)\cos\!\bigl(\tfrac{x}{2}\bigr) \;\Longrightarrow\; \sin^2 x = 4\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)\cos^2\!\bigl(\tfrac{x}{2}\bigr).
Então
\frac{1}{2\,\sin^2\!\bigl(\tfrac{x}{2}\bigr)\cos^2\!\bigl(\tfrac{x}{2}\bigr)} = \frac{2}{\sin^2 x}.
Logo a expressão final é
\frac{2}{\sin^2 x} \;-\; 1 = \frac{2 - \sin^2 x}{\sin^2 x}.

Resposta

\frac{2 - \sin^2 x}{\sin^2 x}.

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