Question

If $\tan \alpha =\frac{1}{7}$, $\tan \beta =\frac{1}{\sqrt{10}}$, prove that $\alpha +2\beta =\frac{\pi }{a}$, where $0<\alpha <\frac{\pi }{2}$ and $0<\beta <\frac{\pi }{2}$

Find $a$.

Answer 1

$\tan (\alpha +2\beta )=\frac{\tan \alpha +\tan 2\beta }{1-\tan \alpha \tan 2\beta }=\frac{\frac{1}{7}+\tan 2\beta }{1-\frac{1}{7}\tan 2\beta }$ (i)

Now, $\tan 2\beta =\frac{2\tan \beta }{1-{\tan }^2\beta }=\frac{2\times \frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}$ [$\tan \beta >0$ as $0<\beta <\pi /2$]

Substituting the value of $\tan 2\beta$ in Eq. (i), we get

$\tan (\alpha +2\beta )=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}\times \frac{3}{4}}=\frac{25}{25}=1$

Now, $0<\alpha <\frac{\pi }{2}$ and $0<\beta <\frac{\pi }{2}$

$\therefore$ $0<\beta <\frac{\pi }{2}$, but $\tan 2\beta =\frac{3}{4}>0$

$\Rightarrow$ $0<2\beta <\frac{\pi }{2}$

Hence, $0<\alpha +2\beta <\pi$

In the interval $(0,\pi )$, $\tan \theta$ takes value 1 at $\pi /4$ only. Therefore,

$\tan (\alpha +2\beta )=tan\left(\frac{\pi }{4}\right)$

$\alpha +2\beta =\frac{\pi }{4}$

Comparing with $\alpha +2\beta =\frac{\pi }{a}$
$\therefore a=4$