Proving Trigonometry Identities $\tan(\frac{1}{2}x+45)+\cot(\frac{1}{2}x+45)=2\sec x$

Question

How do i prove that

$\tan(\frac{1}{2}x+45)+\cot(\frac{1}{2}x+45)=2\sec x$?

I managed to arrive at

$$\frac{2 \sec^2 (\frac{1}{2}x)}{1- \tan^2 (\frac{1}{2}x)}$$

but I got stuck here. Any help is appreciated.

Answer 1

Because $$\tan\left(\frac{x}{2}+45^{\circ}\right)+\cot\left(\frac{x}{2}+45^{\circ}\right)=\frac{1}{\sin\left(\frac{x}{2}+45^{\circ}\right)\cos\left(\frac{x}{2}+45^{\circ}\right)}=\frac{2}{\sin(x+90^{\circ})}=\frac{2}{\cos{x}}.$$

Answer 2

\begin{align*} \tan\left(\frac{1}{2}x+45^{\circ}\right)+\cot\left(\frac{1}{2}x+45^{\circ}\right)&=\tan\left(\frac{1}{2}x+45^{\circ}\right)+\frac1{\tan\left(\frac{1}{2}x+45^{\circ}\right)}\\ &=\frac{1+\tan^2(\frac{1}{2}x+45^{\circ})}{\tan(\frac{1}{2}x+45^{\circ})}\\ &=\frac{\sec^2(\frac{1}{2}x+45^{\circ})}{\tan(\frac{1}{2}x+45^{\circ})}\\ &=\frac{\sec(\frac{1}{2}x+45^{\circ})}{\sin(\frac{1}{2}x+45^{\circ})}\\ &=\frac{2}{2\sin(\frac{1}{2}x+45^{\circ})\cos(\frac{1}{2}x+45^{\circ})}\\ &=\frac{2}{\sin(x+90^{\circ})}\\ &=\frac{2}{\cos(x)}\\ &=2\sec x \end{align*}