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

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

Aicharya · Accepted Answer

Given, $\alpha=\beta+\gamma$ $\begin{aligned} & 	herefore \quad \gamma=\alpha-\beta \ & \quad \begin{aligned} 	an \gamma & =	an (\alpha-\beta) \ & =\frac{	an \alpha-	an \beta}{1+	an \alpha \cdot 	an \beta} \end{aligned} \end{aligned}$ $\begin{aligned} & =\frac{	an \alpha-	an \beta}{1+	an \alpha \cdot 	an \left(\frac{\pi}{2}-\alphaight)} \quad \cdots\left[\begin{array}{l} \alpha+\beta=\frac{\pi}{2} \ 	herefore \beta=\frac{\pi}{2}-\alpha \end{array}ight] \ & =\frac{	an \alpha-	an \beta}{1+	an \alpha \cot \alpha} \ & =\frac{	an \alpha-	an \beta}{2} \end{aligned}$ $\begin{aligned} & \Rightarrow 2 	an \gamma=	an \alpha-	an \beta \ & \Rightarrow 	an \alpha=2 	an \gamma+	an \beta \end{aligned}$

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

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