## Trigonometry (11th Edition) Clone

$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$ The proof is below in the work step by step.
$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$ We examine the right side first. $$A=\frac{2\tan\theta}{1+\tan^2\theta}$$ As from quotient identity: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ so, $$A=\frac{\frac{2\sin\theta}{\cos\theta}}{1+\frac{\sin^2\theta}{\cos^2\theta}}$$ $$A=\frac{\frac{2\sin\theta}{\cos\theta}}{\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}}$$ Now recall that $\cos^2\theta+\sin^2\theta=1$. Thus, $$A=\frac{\frac{2\sin\theta}{\cos\theta}}{\frac{1}{\cos^2\theta}}$$ $$A=\frac{2\sin\theta\cos^2\theta}{\cos\theta\times1}$$ $$A=2\sin\theta\cos\theta$$ Finally, we can rewrite $2\sin\theta\cos\theta$ into $\sin2\theta$, according to double-angle identity for sines. $$A=\sin2\theta$$ Therefore, $$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}$$ The equation is an identity.