If alpha + beta = frac{pi}{2} and beta + gamm | vedclass

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(C) Given $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$,so $	an \beta = \cot \alpha = \frac{1}{	an \alpha}$.Also,$\beta

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$	an \beta + 2	an \gamma$

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(C) Given $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$,so $	an \beta = \cot \alpha = \frac{1}{	an \alpha}$.Also,$\beta + \gamma = \alpha$,so $	an(\beta + \gamma) = 	an \alpha$.Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get $	an \alpha = \frac{	an \beta + 	an \gamma}{1 - 	an \beta 	an \gamma}$.Substituting $	an \beta = \cot \alpha$:$	an \alpha = \frac{\cot \alpha + 	an \gamma}{1 - \cot \alpha 	an \gamma}$.Multiply both sides by $(1 - \cot \alpha 	an \gamma)$:$	an \alpha (1 - \cot \alpha 	an \gamma) = \cot \alpha + 	an \gamma$.$	an \alpha - (	an \alpha \cot \alpha) 	an \gamma = \cot \alpha + 	an \gamma$.Since $	an \alpha \cot \alpha = 1$,we have $	an \alpha - 	an \gamma = \cot \alpha + 	an \gamma$.Since $\cot \alpha = 	an \beta$,we get $	an \alpha = 	an \beta + 2	an \gamma$.

If $\alpha + \beta = \frac{\pi}{2}$ and $\beta + \gamma = \alpha$,then $\tan \alpha$ equals

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