The correct option is B tan α/2+β/2+γ/2=tanα/2tanβ/2tanγ/2 #160;α +β +γ =2π⇒( α +β +γ #160;/ 2 #160;) =π #160;⇒α #160;/ 2 +β #160;/ 2 ...

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Integration by Partial Fractions

If α+β+γ=2π...

Question

If $α + β + γ = 2 π$ , then

t a n \frac{α}{2} + \frac{β}{2} + \frac{γ}{2} = t a n \frac{α}{2} t a n \frac{β}{2} t a n \frac{γ}{2}

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t a n \frac{α}{2} + \frac{β}{2} + \frac{β}{2} = t a n \frac{γ}{2} t a n \frac{γ}{2} t a n \frac{α}{2} = 1

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t a n \frac{α}{2} + \frac{β}{2} + \frac{γ}{2} = - t a n \frac{α}{2} t a n \frac{β}{2} t a n \frac{γ}{2}

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t a n \frac{α}{2} + \frac{β}{2} + \frac{β}{2} = t a n \frac{γ}{2} t a n \frac{γ}{2} t a n \frac{α}{2} = 0

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Solution

The correct option is B $t a n \frac{α}{2} + \frac{β}{2} + \frac{γ}{2} = t a n \frac{α}{2} t a n \frac{β}{2} t a n \frac{γ}{2}$
$α + β + γ = 2 π \Rightarrow (\frac{α + β + γ}{2}) = π$
$\Rightarrow \frac{α}{2} + \frac{β}{2} = π - \frac{γ}{2} \Rightarrow tan (\frac{α}{2} + \frac{β}{2}) = tan (π - \frac{γ}{2})$
$\Rightarrow \frac{tan \frac{α}{2} + tan \frac{β}{2}}{1 - tan \frac{α}{2} tan \frac{β}{2}} = - tan \frac{γ}{2}$
$\Rightarrow tan \frac{α}{2} + tan \frac{β}{2} = - tan \frac{γ}{2} + tan \frac{α}{2} tan \frac{β}{2} tan \frac{γ}{2}$
$\Rightarrow tan \frac{α}{2} + tan \frac{β}{2} + tan \frac{γ}{2} = tan \frac{α}{2} tan \frac{β}{2} tan \frac{γ}{2}$

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