Can someone help me visualize this trigonometry Identity
Prove that :
$$1+\tan A\times\tan \frac{A}{2} = \sec A$$
I got the answer by manipulating the lhs substitution lots of mundane stuff but without a pen and paper and say a magical ability to never be able to visualize text in my mind, I wouldn't have been able to visualize this geometrically and understand how $\sec A$ is achieved. Any suggestions ?
Consider a right triangle with one size equal to 1. AD is the is angle A bisector and DE is perpendicular to hypotenuse. Since ABD and AED are congruent, $AE =1$ and $DE = \tan {\frac{A}{2}}$. You just need to write tangent of angle C in DEC and ABC.
$\tan{C} = \frac{1}{\tan{A}} = \frac{\tan{\frac{A}{2}}}{\sec{A}-1} \Rightarrow \tan{A}\tan{\frac{A}{2}}=\sec{A}-1$.
I hope this helps.