Answer
$1+\tan x\tan\dfrac{x}{2}=\sec x$
Work Step by Step
$1+\tan x\tan\dfrac{x}{2}=\sec x$
Substitute $\tan x$ by $\dfrac{\sin x}{\cos x}$ and $\tan\dfrac{x}{2}$ by $\dfrac{\sin x}{1+\cos x}$ (This is one of the half-angle formulas for the tangent):
$1+\Big(\dfrac{\sin x}{\cos x}\Big)\Big(\dfrac{\sin x}{1+\cos x}\Big)=\sec x$
Evaluate the product on the left side:
$1+\dfrac{\sin^{2}x}{\cos x(1+\cos x)}=\sec x$
Substitute $\sin^{2}x$ by $1-\cos^{2}x$:
$1+\dfrac{1-\cos^{2}x}{\cos x(1+\cos x)}=\sec x$
Factor the numerator of the fraction on the left side and simplify:
$1+\dfrac{(1-\cos x)(1+\cos x)}{\cos x(1+\cos x)}=\sec x$
$1+\dfrac{1-\cos x}{\cos x}=\sec x$
Rewrite the fraction on the left as $\dfrac{1}{\cos x}-\dfrac{\cos x}{\cos x}$ and simplify:
$1+\dfrac{1}{\cos x}-\dfrac{\cos x}{\cos x}=\sec x$
$1+\dfrac{1}{\cos x}-1=\sec x$
$\dfrac{1}{\cos x}=\sec x$
Since $\dfrac{1}{\cos x}=\sec x$, the identity is proved.
$\sec x=\sec x$
