If A+B+C=180^{circ},then the value of tan lef | vedclass

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(A) Given $A+B+C=180^{\circ}$,so $\frac{A+B+C}{2} = 90^{\circ}$.Thus,$\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$.Taking tangent on both sides: $	an \l

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(A) Given $A+B+C=180^{\circ}$,so $\frac{A+B+C}{2} = 90^{\circ}$.Thus,$\frac{A+B}{2} = 90^{\circ} - \frac{C}{2}$.Taking tangent on both sides: $	an \left(\frac{A+B}{2}ight) = 	an \left(90^{\circ} - \frac{C}{2}ight) = \cot \left(\frac{C}{2}ight)$.Using the formula $	an(x+y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we get:$\frac{	an(A/2) + 	an(B/2)}{1 - 	an(A/2) 	an(B/2)} = \frac{1}{	an(C/2)}$.Cross-multiplying gives: $	an(C/2) [	an(A/2) + 	an(B/2)] = 1 - 	an(A/2) 	an(B/2)$.Rearranging the terms: $	an(A/2) 	an(B/2) + 	an(B/2) 	an(C/2) + 	an(C/2) 	an(A/2) = 1$.

If $A+B+C=180^{\circ}$,then the value of $\tan \left(\frac{A}{2}\right) \tan \left(\frac{B}{2}\right)+\tan \left(\frac{B}{2}\right) \tan \left(\frac{C}{2}\right)+\tan \left(\frac{C}{2}\right) \tan \left(\frac{A}{2}\right)$ is

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