Prove: $\tan{\frac{x}{2}}\sec{x}= \tan{x} - \tan{\frac{x}{2}}$

Question

I was solving a question which required the above identity to proceed but I never found its proof anywhere. I tried to prove it but got stuck after a while.

I reached till here:

To Prove: $$\tan{\frac{x}{2}}\sec{x}= \tan{x} - \tan{\frac{x}{2}}$$

But I don't know what to do next. Any help is appreciated Thanks

Answer 1

$$\tan \dfrac x2\sec x=\dfrac{\sin\dfrac x2}{\cos\dfrac x2\cdot\cos x}$$

Use $\sin\dfrac x2=\sin\left(x-\dfrac x2\right)=?$

Answer 2

Hint:

use Weierstrass substitution

$$\tan\dfrac x2\sec x+\tan\dfrac x2=\tan\dfrac x2\left(\dfrac{1+\tan^2\dfrac x2}{1-\tan^2\dfrac x2}+1\right)=?$$

Answer 3

In terms of $t=\tan\tfrac{x}{2}$, the LHS is $t\cdot\frac{1+t^2}{1-t^2}$, while the RHS is $\frac{2t}{1-t^2}-t=t\cdot\frac{2-(1-t^2)}{1-t^2}=t\cdot\frac{1+t^2}{1-t^2}$.

Answer 4

For fun, here's a trigonographic solution:

$$|\triangle ORS| = |\triangle OTS| - |\triangle OTR| \;\to\; \color{gray}{\tfrac12}\cdot\tan\theta\cdot\sec2\theta=\color{gray}{\tfrac12}\cdot 1\cdot\tan2\theta-\color{gray}{\tfrac12}\cdot1\cdot\tan\theta$$