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Question

(C) Given $\alpha + \beta = \pi \implies \alpha = \pi - \beta \implies 	an \alpha = -	an \beta$.Also,$\beta + \gamma = \alpha$.Taking tangent on bot

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given $\alpha + \beta = \pi \implies \alpha = \pi - \beta \implies 	an \alpha = -	an \beta$.Also,$\beta + \gamma = \alpha$.Taking tangent on both sides: $	an(\beta + \gamma) = 	an \alpha$.$\frac{	an \beta + 	an \gamma}{1 - 	an \beta 	an \gamma} = 	an \alpha$.Substitute $	an \beta = -	an \alpha$:$\frac{-	an \alpha + 	an \gamma}{1 - (-	an \alpha) 	an \gamma} = 	an \alpha$.$-	an \alpha + 	an \gamma = 	an \alpha(1 + 	an \alpha 	an \gamma) = 	an \alpha + 	an^2 \alpha 	an \gamma$.$	an^2 \alpha 	an \gamma = -2 	an \alpha + 	an \gamma$.Since $	an \alpha = -	an \beta$,we have $	an^2 \alpha = 	an \alpha \cdot (-	an \beta) = -	an \alpha 	an \beta$.Substituting $	an \alpha = -	an \beta$ into the equation $	an^2 \alpha 	an \gamma = -2 	an \alpha + 	an \gamma$:$	an^2 \alpha 	an \gamma = 2 	an \beta + 	an \gamma$.$	an^2 \alpha = \frac{2 	an \beta + 	an \gamma}{	an \gamma}$.Since $\alpha = \pi - \beta$ and $\beta, \gamma$ are positive,$\alpha$ is in the second quadrant,so $	an \alpha$ must be negative.Therefore,$	an \alpha = -\sqrt{\frac{2 	an \beta + 	an \gamma}{	an \gamma}}$.

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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