#### Answer

$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$
The equation is an identity, as verified below.

#### Work Step by Step

$$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$
1) We start from the left side.
$$X=\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)$$
- Recall the sum identity for tangent, which states
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
Replace the identity into $X$ with $A=\frac{x}{2}$ and $B=\frac{\pi}{4}$, we have
$$X=\frac{\tan\frac{x}{2}+\tan\frac{\pi}{4}}{1-\tan\frac{x}{2}\tan\frac{\pi}{4}}$$
$$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}\times1}$$
$$X=\frac{\tan\frac{x}{2}+1}{1-\tan\frac{x}{2}}$$
- Now use the half-angle identity for tangent: $$\tan\frac{x}{2}=\frac{1-\cos x}{\sin x}$$
Thus, $$X=\frac{\frac{1-\cos x}{\sin x}+1}{1-\frac{1-\cos x}{\sin x}}$$
$$X=\frac{\frac{1-\cos x+\sin x}{\sin x}}{\frac{\sin x-1+\cos x}{\sin x}}$$
$$X=\frac{1-\cos x+\sin x}{\sin x-1+\cos x}$$
$$X=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$
We stop here as there seems no potential in simplifying anymore.
2) Then we look at the right side.
$$Y=\sec x+\tan x$$
- Apply the following identities:
$$\sec x=\frac{1}{\cos x}\hspace{2cm}\tan x=\frac{\sin x}{\cos x}$$
$$Y=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$$
$$Y=\frac{1+\sin x}{\cos x}$$
So the job here is to prove that $\frac{1+\sin x}{\cos x}=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$, which seems so different.
So in a seemingly dead-end case like this, we would try multiplying both the numerator and denominator of $Y$ with $\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$.
$$Y=\frac{1+\sin x}{\cos x}\times\frac{\sin x+\cos x-1}{\sin x+\cos x-1}$$
$$Y=\frac{\sin x+\cos x-1+\sin^2x+\sin x\cos x-\sin x}{\cos x(\sin x+\cos x-1)}$$
$$Y=\frac{\sin^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$
- Now remember that $\sin^2x=1-\cos^2x$, which can be used to rewrite $\sin^2x$.
$$Y=\frac{1-\cos^2x+\sin x\cos x+\cos x-1}{\cos x(\sin x+\cos x-1)}$$
$$Y=\frac{-\cos^2x+\sin x\cos x+\cos x}{\cos x(\sin x+\cos x-1)}$$
$$Y=\frac{\cos x(\sin x-\cos x+1)}{\cos x(\sin x+\cos x-1)}$$
$$Y=\frac{\sin x-\cos x+1}{\sin x+\cos x-1}$$
So after a rather painstaking process of transformations, we show that $X=Y$.
Therefore, $$\tan\Big(\frac{x}{2}+\frac{\pi}{4}\Big)=\sec x+\tan x$$
The equation is therefore an identity.