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(C) Given $\alpha + \beta = \pi$ and $\beta + \gamma = \alpha$.From $\alpha + \beta = \pi$,we have $\alpha = \pi - \beta$,so $	an \alpha = 	an(\pi -

Vedclass · Accepted Answer

$-\sqrt{\frac{2	an \beta + 	an \gamma}{	an \gamma}}$

Vedclass · Answer

(C) Given $\alpha + \beta = \pi$ and $\beta + \gamma = \alpha$.From $\alpha + \beta = \pi$,we have $\alpha = \pi - \beta$,so $	an \alpha = 	an(\pi - \beta) = -	an \beta$.Since $\alpha, \beta, \gamma$ are positive and $\alpha + \beta = \pi$,$\alpha$ must be in the second quadrant (as $\beta > 0$),so $	an \alpha From $\beta + \gamma = \alpha$,we have $	an \alpha = 	an(\beta + \gamma) = \frac{	an \beta + 	an \gamma}{1 - 	an \beta 	an \gamma}$.Substituting $	an \beta = -	an \alpha$ into the equation:$	an \alpha = \frac{-	an \alpha + 	an \gamma}{1 - (-	an \alpha) 	an \gamma} = \frac{-	an \alpha + 	an \gamma}{1 + 	an \alpha 	an \gamma}$.$	an \alpha (1 + 	an \alpha 	an \gamma) = -	an \alpha + 	an \gamma$.$	an \alpha + 	an^2 \alpha 	an \gamma = -	an \alpha + 	an \gamma$.$	an^2 \alpha 	an \gamma = -2 	an \alpha + 	an \gamma$.This is a quadratic in $	an \alpha$: $(	an \gamma) 	an^2 \alpha + (2) 	an \alpha - 	an \gamma = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$	an \alpha = \frac{-2 \pm \sqrt{4 - 4(	an \gamma)(-	an \gamma)}}{2 	an \gamma} = \frac{-2 \pm \sqrt{4 + 4 	an^2 \gamma}}{2 	an \gamma} = \frac{-2 \pm 2 \sqrt{1 + 	an^2 \gamma}}{2 	an \gamma} = \frac{-1 \pm \sec \gamma}{	an \gamma}$.Alternatively,from $	an^2 \alpha 	an \gamma = -2 	an \alpha + 	an \gamma$,we get $	an^2 \alpha = \frac{-2 	an \alpha + 	an \gamma}{	an \gamma}$. Since $	an \alpha = -	an \beta$,$	an^2 \alpha = \frac{2 	an \beta + 	an \gamma}{	an \gamma}$.Thus,$	an \alpha = -\sqrt{\frac{2 	an \beta + 	an \gamma}{	an \gamma}}$ (taking the negative root because $	an \alpha < 0$).

If $\alpha, \beta, \gamma$ are positive numbers such that $\alpha + \beta = \pi$ and $\beta + \gamma = \alpha$,then $\tan \alpha$ is equal to - (where $\gamma \neq n\pi, n \in I$)

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