MCQ
If $\tan \alpha = \frac{m}{{m + 1}}$ and $\tan \beta = \frac{1}{{2m + 1}}$, then $\alpha + \beta = $
- A$\frac{\pi }{3}$
- ✓$\frac{\pi }{4}$
- C$\frac{\pi }{6}$
- DNone of these
and $\tan \,\beta = \frac{1}{{2m + 1}}$
We know $\tan \,(\alpha + \beta ) = \frac{{\tan \,\alpha + \tan \,\beta }}{{1 - \tan \,\alpha \,\tan \,\beta }}$
$ = \frac{{\frac{m}{{m + 1}} + \frac{1}{{2m + 1}}}}{{1 - \frac{m}{{(m + 1)}}\,\frac{1}{{(2m + 1)}}}} = \frac{{2{m^2} + m + m + 1}}{{2{m^2} + m + 2m + 1 - m}}$
$ = \frac{{2{m^2} + 2m + 1}}{{2{m^2} + 2m + 1}} = 1\,\,$
$\Rightarrow \,\,\tan \,(\alpha + \beta ) = \tan \frac{\pi }{4}$
Hence, $\alpha + \beta = \frac{\pi }{4}$.
Trick : As $\alpha + \beta $ is independent of $m$,
therefore put $m = 1,$ તો $\tan \,\alpha = \frac{1}{2}$ and $\tan \,\beta = \frac{1}{3}$.
Therefore, $\tan \,(\alpha + \beta ) = \frac{{(1/2) + (1/3)}}{{1 - (1/6)}} = 1.$
Hence $\alpha + \beta = \frac{\pi }{4}.$
(Also check for other values of $m$).
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$2\left(a_1+a_2+\ldots .+a_n\right)=b_1+b_2+\ldots . .+b_n$
holds for some positive integer $n$, is. . . . . . .