$\Rightarrow \tan \beta = \frac{1}{3}$
==> $\tan 2\beta = \frac{{2\tan \beta }}{{1 - {{\tan }^2}\beta }} = \frac{3}{4}$
$\therefore \tan (\alpha + 2\beta ) = \frac{{\frac{1}{7} + \frac{3}{4}}}{{1 - \frac{1}{7}.\frac{3}{4}}} = \frac{{25}}{{25}} = 1$
Now, $0 < \beta < \frac{\pi }{2}$ and $\tan 2\beta = \frac{3}{4} > 0$ both
==> $0 < 2\beta < \frac{\pi }{2}$.
Again,$0 < \alpha < \frac{\pi }{2}$ and $0 < 2\beta < \frac{\pi }{2}$ both
==> $0 < \alpha + 2\beta < \pi $
Thus, $0 < \alpha + 2\beta < \pi $ and $\tan (\alpha + 2\beta ) = 1$ both
==> $\alpha + 2\beta = \frac{\pi }{4} $
$\Rightarrow 2\beta = \frac{\pi }{4} - \alpha $.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| $x_i$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
| $f_i$ | $4$ | $4$ | $\alpha$ | $15$ | $8$ | $\beta$ | $4$ | $5$ |
are $9$ and $15.08$ respectively, then the value of $\alpha^2+\beta^2-\alpha \beta$ is $............$.