3. trignometrical ratios,functions and identities — ગણિત ધોરણ 11 સાયન્સ — Question

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 $m$).