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

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

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  • $\begingroup$ There is a useful identity $$\tan\left(\dfrac{\theta}{2}\right)=\frac{\sin(\theta)}{1+\cos(\theta)}=\frac{1-\cos(\theta)}{\sin(\theta)}\,.$$ $\endgroup$ – Batominovski Jul 6 '20 at 14:00
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$$\tan \dfrac x2\sec x=\dfrac{\sin\dfrac x2}{\cos\dfrac x2\cdot\cos x}$$

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

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For fun, here's a trigonographic solution:

enter image description here

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

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    $\begingroup$ That was really fun to solve! I never proved any identity by geometry except the basic ones. $\endgroup$ – Kaustubh Gupta Jul 6 '20 at 14:33
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    $\begingroup$ @KaustubhGupta: I'm glad you like it. :) I have a number of these kinds of diagrams on trigonography.com. I've neglected that site for a while, but I think the above figure would make a nice addition. Thanks for bringing that identity to my attention. $\endgroup$ – Blue Jul 6 '20 at 14:39
  • $\begingroup$ That would be really nice to add. Glad I could help.:D $\endgroup$ – Kaustubh Gupta Jul 6 '20 at 14:45
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Hint:

use Weierstrass substitution

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

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

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