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Question Number 43874 by peter frank last updated on 16/Sep/18

$$in\mathrm{\Delta }ABC,provethat$$$$\frac{a+b-c}{a+b+c}=\tan \frac{A}{2}\tan \frac{B}{2}$$

Answered by ajfour last updated on 16/Sep/18

$$\frac{a+b-c}{a+b+c}=\frac{\sin A+\sin B-\sin (A+B)}{\sin A+\sin B+\sin (A+B)}$$$$=\frac{2\sin \left(\frac{A+B}{2}\right)[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)]}{2\sin \left(\frac{A+B}{2}\right)[\cos \left(\frac{A-B}{2}\right)+\cos \left(\frac{A+B}{2}\right)]}$$$$=\frac{2\sin \frac{A}{2}\sin \frac{B}{2}}{2\cos \frac{A}{2}\cos \frac{B}{2}}=\tan \frac{A}{2}\tan \frac{B}{2}.$$

Commented by peter frank last updated on 17/Sep/18

$$thankyou$$$$$$

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