In Delta ABC, 1 - tan frac{A}{2}tan frac{B}{2 | vedclass

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(A) We know that in $\Delta ABC,$ $A + B + C = 180^{\circ},$ so $\frac{A}{2} + \frac{B}{2} = 90^{\circ} - \frac{C}{2}.$$1 - 	an \frac{A}{2}	an \frac

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$\frac{2c}{a + b + c}$

Vedclass · Answer

(A) We know that in $\Delta ABC,$ $A + B + C = 180^{\circ},$ so $\frac{A}{2} + \frac{B}{2} = 90^{\circ} - \frac{C}{2}.$$1 - 	an \frac{A}{2}	an \frac{B}{2} = \frac{\cos \frac{A}{2}\cos \frac{B}{2} - \sin \frac{A}{2}\sin \frac{B}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}$$= \frac{\cos(\frac{A}{2} + \frac{B}{2})}{\cos \frac{A}{2}\cos \frac{B}{2}} = \frac{\cos(90^{\circ} - \frac{C}{2})}{\cos \frac{A}{2}\cos \frac{B}{2}} = \frac{\sin \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{B}{2}}$Using the half-angle formulas $\sin \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}},$ $\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}},$ and $\cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}},$ we get:$= \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{s(s-a)}{bc}} \cdot \sqrt{\frac{s(s-b)}{ac}}} = \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{s^2(s-a)(s-b)}{abc^2}}} = \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\frac{s}{c} \sqrt{\frac{(s-a)(s-b)}{ab}}} = \frac{c}{s}$Since $s = \frac{a+b+c}{2},$ we have $\frac{c}{s} = \frac{c}{(a+b+c)/2} = \frac{2c}{a+b+c}.$

In $\Delta ABC,$ $1 - \tan \frac{A}{2}\tan \frac{B}{2} = $

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