1. We need to prove the identity $$\frac{1+\sin(2x)}{\cos(2x)} = \tan\left(\frac{\pi}{4} x\right)$$. 2. Recall the double-angle formulas: - $\sin(2x) = 2\sin x \cos x$ - $\cos(2x) = \cos^2 x - \sin^2 x$ 3. Start with the left side (LHS): $$\frac{1+\sin(2x)}{\cos(2x)}$$ 4. Substitute $\sin(2x)$: $$\frac{1 + 2\sin x \cos x}{\cos(2x)}$$ 5. Use the identity $\cos(2x) = 1 - 2\sin^2 x$ (one form of cosine double angle): $$\frac{1 + 2\sin x \cos x}{1 - 2\sin^2 x}$$ 6. Let $t = \tan\left(\frac{\pi}{4} x\right)$, we want to show LHS = $t$. 7. Note that $\tan\left(\frac{\pi}{4} x\right)$ is not a standard double angle formula, so let's check if the identity holds for specific values or if there is a typo. 8. Testing $x=0$: LHS: $\frac{1 + \sin 0}{\cos 0} = \frac{1+0}{1} = 1$ RHS: $\tan(0) = 0$ Not equal. 9. Testing $x=\frac{\pi}{2}$: LHS: $\frac{1 + \sin(\pi)}{\cos(\pi)} = \frac{1+0}{-1} = -1$ RHS: $\tan\left(\frac{\pi}{4} \cdot \frac{\pi}{2}\right) = \tan\left(\frac{\pi^2}{8}\right)$ which is not $-1$. 10. Since the identity does not hold for these values, it is false as stated. Therefore, the given identity $$\frac{1+\sin(2x)}{\cos(2x)} = \tan\left(\frac{\pi}{4} x\right)$$ is not true in general.