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

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(C) Given,$\alpha = \beta + \gamma$. Since $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$. Substituting $\beta$ in the firs

Vedclass · Accepted Answer

$	an \beta+2 	an \gamma$

Vedclass · Answer

(C) Given,$\alpha = \beta + \gamma$. Since $\alpha + \beta = \frac{\pi}{2}$,we have $\beta = \frac{\pi}{2} - \alpha$. Substituting $\beta$ in the first equation: $\alpha = (\frac{\pi}{2} - \alpha) + \gamma$,which implies $\gamma = 2\alpha - \frac{\pi}{2}$. Alternatively,using $\gamma = \alpha - \beta$: $	an \gamma = 	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$. Since $\beta = \frac{\pi}{2} - \alpha$,then $	an \beta = \cot \alpha = \frac{1}{	an \alpha}$. Substituting this: $	an \gamma = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha (\frac{1}{	an \alpha})} = \frac{	an \alpha - 	an \beta}{1 + 1} = \frac{	an \alpha - 	an \beta}{2}$. Therefore,$2 	an \gamma = 	an \alpha - 	an \beta$,which gives $	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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