If A+B+C=π and sin(A+C/2)=n sin C/2,Show that tan A/2 tan B/2=n-1/n+1

If A+B+C=π and sin(A+C/2)=n sin C/2,Show that tan A/2 tan B/2=n-1/n+1

If A+B+C=π and sin(A+C/2)=n sin C/2,Show that tan A/2 tan B/2=n-1/n+1

$\sin (A + \frac{C}{2})$= $nsin\frac{C}{2}$

or,$\frac{\sin (A + \frac{C}{2})}{sin\frac{C}{2}}$= $\frac{n}{1}$

or,$\frac{\sin (A + \frac{C}{2}) + sin\frac{C}{2}}{\sin (A + \frac{C}{2}) – sin\frac{C}{2}}$=$\frac{n + 1}{n – 1}$

or,$\frac{2sin\frac{A + \frac{C}{2} + \frac{C}{2}}{2}\cos \frac{n + \frac{C}{2} – \frac{C}{2}}{2}}{2cos\frac{A + \frac{C}{2} + \frac{C}{2}}{2}sin\frac{A + \frac{C}{2} – \frac{C}{2}}{2}}$ =$\frac{n + 1}{n – 1}$

or,$\frac{2sin\frac{A + C}{2}\cos \frac{A}{2}}{2cos\frac{A + C}{2}\sin \frac{A}{2}}$ = $\frac{n + 1}{n – 1}$

or,$tan\frac{A + C}{2}cot\frac{A}{2}$ = $\frac{n + 1}{n – 1}$

${\mathrm{or}},tan\frac{\pi – B}{2}\cot \frac{A}{2}$ = $\frac{n + 1}{n – 1}$ [∵A+B+C=π]

${\mathrm{or}},tan(\frac{{\mathrm{\pi}}}{2} – \frac{{\mathrm{B}}}{2})cot\frac{A}{2}$ = $\frac{n + 1}{n – 1}$

${\mathrm{or}},c{\mathrm{ot}}\frac{{\mathrm{B}}}{2}\cot \frac{{\mathrm{A}}}{2}$ = $\frac{n + 1}{n – 1}$

${\mathrm{or}},tan\frac{{\mathrm{A}}}{2}tan\frac{{\mathrm{B}}}{2}$ = $\frac{{\mathrm{n}} – 1}{{\mathrm{n}} + 1}$

$\therefore tan\frac{{\mathrm{A}}}{2}tan\frac{{\mathrm{B}}}{2}$ = $\frac{{\mathrm{n}} – 1}{{\mathrm{n}} + 1}$[Proved]