If alpha + beta + gamma = 2pi , then | vedclass

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(A) Given $\alpha + \beta + \gamma = 2\pi$. Dividing by $2$,we get $\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi$. Taking tangent on bo

Vedclass · Accepted Answer

$	an \frac{\alpha }{2} + 	an \frac{\beta }{2} + 	an \frac{\gamma }{2} = 	an \frac{\alpha }{2}	an \frac{\beta }{2}	an \frac{\gamma }{2}$

Vedclass · Answer

(A) Given $\alpha + \beta + \gamma = 2\pi$. Dividing by $2$,we get $\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi$. Taking tangent on both sides,$	an(\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2}) = 	an(\pi) = 0$. Using the identity $	an(A+B+C) = \frac{\sum 	an A - \prod 	an A}{1 - \sum 	an A 	an B}$,we have: $\frac{	an \frac{\alpha}{2} + 	an \frac{\beta}{2} + 	an \frac{\gamma}{2} - 	an \frac{\alpha}{2}	an \frac{\beta}{2}	an \frac{\gamma}{2}}{1 - (	an \frac{\alpha}{2}	an \frac{\beta}{2} + 	an \frac{\beta}{2}	an \frac{\gamma}{2} + 	an \frac{\gamma}{2}	an \frac{\alpha}{2})} = 0$. Since the denominator is not zero,the numerator must be zero: $	an \frac{\alpha}{2} + 	an \frac{\beta}{2} + 	an \frac{\gamma}{2} - 	an \frac{\alpha}{2}	an \frac{\beta}{2}	an \frac{\gamma}{2} = 0$. Therefore,$	an \frac{\alpha}{2} + 	an \frac{\beta}{2} + 	an \frac{\gamma}{2} = 	an \frac{\alpha}{2}	an \frac{\beta}{2}	an \frac{\gamma}{2}$.

If $\alpha + \beta + \gamma = 2\pi ,$ then

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