Question

simplify sin(-theta)/tan(-theta)

Answer 1

\[\frac{ \sin(-\theta) }{ \tan(-\theta) }\]

Answer 2

\[\frac{ \sin (-\theta) }{ \frac{ \sin(-\theta) }{ \cos (\theta) }}\]

Answer 3

Recall \(\sin\) is even so \(\sin-\theta=-\sin\theta\). Similarly, \(\tan\) is also even so \(\tan-\theta=-\tan\theta\). We're left with:$$\frac{\sin-\theta}{\tan-\theta}=\frac{-\sin\theta}{-\tan\theta}=\frac{\sin\theta}{\tan\theta}$$We're not done yet, though. Recall that \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\):$$\frac{\sin\theta}{\tan\theta}=\sin\theta\cdot\frac1{\tan\theta}=\sin\theta\cdot\frac{\cos\theta}{\sin\theta}=\cos\theta$$and we're done.

Answer 4

Thank you!