Respuesta :
Given the expression:
[tex]\csc \theta-\cot \theta\cos (-\theta)[/tex]Let's write the expression in terms of sinθ o cosθ.
To simplify, apply trigonometric identities.
[tex]\begin{gathered} \csc (\theta)=\frac{1}{\sin\theta} \\ \\ \text{cot}\theta=\frac{\cos \theta}{\sin \theta} \end{gathered}[/tex]Thus, we have:
[tex]\begin{gathered} \csc \theta-\cot \theta\cos (-\theta) \\ \\ =\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}\cdot\cos (-\theta) \end{gathered}[/tex]Solving further:
Since cos(-θ) is an even function, rewrite it as cos(θ)
[tex]\begin{gathered} \frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}\cdot\cos \theta \\ \\ \frac{1}{\sin\theta}-\frac{\cos \theta\cdot\cos \theta}{\sin \theta} \\ \\ \frac{1}{\sin\theta}-\frac{\cos ^2\theta}{\sin \theta} \\ \\ =\frac{1-\cos ^2\theta}{\sin \theta} \end{gathered}[/tex]Apply Pythagorean Identity:
[tex]\begin{gathered} \frac{\sin^2\theta}{\sin\theta} \\ \\ =\frac{\sin \theta\cdot\sin \theta}{\sin \theta} \\ \\ =\sin \theta \end{gathered}[/tex]ANSWER:
[tex]\sin \theta[/tex]