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

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(A) Given that $\alpha + \beta + \gamma = 2\pi$.Dividing by $2$,we get $\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi$.Taking the tangen

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 that $\alpha + \beta + \gamma = 2\pi$.Dividing by $2$,we get $\frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi$.Taking the tangent of both sides,we have $	an \left( \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} ight) = 	an \pi = 0$.Using the formula $	an(A+B+C) = \frac{\sum 	an A - \prod 	an A}{1 - \sum 	an A 	an B}$,we get:$\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 fraction is $0$,the numerator must be $0$:$	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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