If $ an{\alpha = \frac{m}{m + 1}}$ and $ an{\beta = \frac{1}{2m + 1}}$, then $\alpha + \beta$ is equal to

Question

If $	an{\alpha = \frac{m}{m + 1}}$ and $	an{\beta = \frac{1}{2m + 1}}$, then $\alpha + \beta$ is equal to

Dr. Abubacker · Accepted Answer

(b)
$	an{(\alpha + \beta) = \frac{	an{\alpha + 	an\beta}}{1 - 	an{\alpha	an\beta}}}$
$= \frac{\frac{m}{m + 1} + \frac{1}{2m + 1}}{1 - \frac{m}{m + 1}.\frac{1}{2m + 1}}$
$= \frac{2m^{2} + m + m + 1}{2m^{2} + 3m + 1 - m} = 1 = 	an\frac{\pi}{4}$
$	herefore\ \ \ \alpha + \beta = \frac{\pi}{4}$

Q. If $\tan{\alpha = \frac{m}{m + 1}}$ and $\tan{\beta = \frac{1}{2m + 1}}$, then $\alpha + \beta$ is equal to

Explanation

Reviewed by Dr. Abubacker

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