MCQ
$(m + 2)\sin \theta + (2m - 1)\cos \theta = 2m + 1,$ જો . . .
- A$\tan \theta = \frac{3}{4}$
- ✓$\tan \theta = \frac{4}{3}$
- C$\tan \theta = \frac{{2m}}{{{m^2} + 1}}$
- Dએકપણ નહિ.
${(m + 2)^2}\,{t^2} + 2(m + 2)\,(2m - 1)t + {(2m - 1)^2} = {(2m + 1)^2}\,(1 + {t^2})$
$ \Rightarrow \,3\,(1 - {m^2})\,{t^2} + (4{m^2} + 6m - 4)\,t - 8m = 0$
$ \Rightarrow \,(3t - 4)\,[(1 - {m^2})\,t + 2m] = 0$,
which is true, if $t = \tan \theta = \frac{4}{3}$ or $\tan \theta = \frac{{2m}}{{{m^2} - 1}}$.
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