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 = 	an(\frac{\pi}{2} - \alpha) = \cot \alpha$.Also,

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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 = 	an(\frac{\pi}{2} - \alpha) = \cot \alpha$.Also,given $\beta + \gamma = \alpha$,taking $	an$ on both sides gives $	an(\beta + \gamma) = 	an \alpha$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get $\frac{	an \beta + 	an \gamma}{1 - 	an \beta 	an \gamma} = 	an \alpha$.Substituting $	an \beta = \cot \alpha = \frac{1}{	an \alpha}$,we have $\frac{\frac{1}{	an \alpha} + 	an \gamma}{1 - \frac{1}{	an \alpha} 	an \gamma} = 	an \alpha$.Multiplying the numerator and denominator by $	an \alpha$,we get $\frac{1 + 	an \alpha 	an \gamma}{	an \alpha - 	an \gamma} = 	an \alpha$.$1 + 	an \alpha 	an \gamma = 	an^2 \alpha - 	an \alpha 	an \gamma$.$1 + 2	an \alpha 	an \gamma = 	an^2 \alpha$.Dividing by $	an \alpha$,we get $\cot \alpha + 2	an \gamma = 	an \alpha$.Since $\cot \alpha = 	an \beta$,we have $	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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