# Help! Cubic trigonometric Identity proof

I was wondering if anybody could help me find the answer to this question. In class we practiced solving trigonometric identities but no cubics. Then this question in one of my practice papers threw me.

$$\frac{\sin\theta-\sin^3\theta}{\cos^3\theta}\equiv \tan\theta$$

TIA for any help

• did you try anything? – Arnaldo Apr 12 '17 at 14:51
• i tried using the basic identities like $\cos^2\theta +\sin^2\theta \equiv 1$ , $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$ but thats all i was tought in class and i cant see how they fit in to this problem – James Bell Apr 12 '17 at 15:00

$$\frac{\sin\theta \left( 1-\sin^2\theta\right)}{\cos^3\theta} = \frac{\sin\theta \cos^2\theta }{\cos^3\theta} = \frac{\sin\theta }{\cos\theta} = \tan\theta$$
• thank you. sorry, can you please explain why we jumped from $\sin\theta (1-\sin^2\theta)$ to $\sin\theta\cos^2\theta$ – James Bell Apr 12 '17 at 14:53
• am i on the right lines? $\frac{\sin\theta -\sin^3\theta}{\cos^3\theta}$ $\cos^2\theta+\sin^2\theta \equiv 1$ so $\cos^2\theta -1 \equiv \sin^2\theta$ so $\frac{\sin\theta-\sin\theta \sin^2\theta}{\cos^3\theta}$ $\frac{\cos^2\theta -1}{\cos^3\theta}\equiv ?$ – James Bell Apr 12 '17 at 15:09