MCQ
If $2\tan A = 3\tan B,$ then $\frac{{\sin 2B}}{{5 - \cos 2B}}$ is equal to
- A$\tan A - \tan B$
- ✓$\tan (A - B)$
- C$\tan (A + B)$
- D$\tan (A + 2B)$
==> $\tan A = \frac{3}{2}\tan B = \frac{3}{2}t$, [Let $\tan B = t$]
==> $\sin 2B = \frac{{2t}}{{1 + {t^2}}},\cos 2B = \frac{{1 - {t^2}}}{{1 + {t^2}}}$
$\therefore$ $\frac{{\left( {\frac{{2t}}{{1 + {t^2}}}} \right)}}{{5 - \left( {\frac{{1 - {t^2}}}{{1 + {t^2}}}} \right)}}$
$ = \frac{{2t}}{{4 + 6{t^2}}} = \frac{t}{{2 + 3{t^2}}} = \tan (A - B)$.
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$(A)$ $\frac{2 \log _3 2}{2 \log _3 2-1}$ $(B)$ $\frac{2}{2-\log _2 3}$ $(C)$ $\frac{1}{1-\log _4 3}$ $(D)$ $\frac{2 \log _2 3}{2 \log _2 3-1}$