## Trigonometry (11th Edition) Clone

$$\frac{\sin\theta+\tan\theta}{1+\cos\theta}=\tan\theta$$ The process of identity verification is shown in the work step by step.
$$\frac{\sin\theta+\tan\theta}{1+\cos\theta}=\tan\theta$$ We examine the left side. $$X=\frac{\sin\theta+\tan\theta}{1+\cos\theta}$$ - First, we rewrite $\tan\theta$ into $\frac{\sin\theta}{\cos\theta}$, as in Quotient Identity. $$X=\frac{\sin\theta+\frac{\sin\theta}{\cos\theta}}{1+\cos\theta}$$ $$X=\frac{\frac{\sin\theta\cos\theta+\sin\theta}{\cos\theta}}{1+\cos\theta}$$ $$X=\frac{\sin\theta\cos\theta+\sin\theta}{\cos\theta(1+\cos\theta)}$$ $$X=\frac{\sin\theta(\cos\theta+1)}{\cos\theta(\cos\theta+1)}$$ $$X=\frac{\sin\theta}{\cos\theta}$$ - Now, in reverse, $\frac{\sin\theta}{\cos\theta}$ can be written into $\tan\theta$. $$X=\tan\theta$$ That means, $$\frac{\sin\theta+\tan\theta}{1+\cos\theta}=\tan\theta$$ The equation is an identity as a result.