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3.Trigonometrical Ratios, Functions and Identities
medium
If $2\tan A = 3\tan B,$ then $\frac{{\sin 2B}}{{5 - \cos 2B}}$ is equal to
A
$\tan A - \tan B$
B
$\tan (A - B)$
C
$\tan (A + B)$
D
$\tan (A + 2B)$
Solution
(b) $2\tan {\rm A} = 3\tan B$
==> $\tan A = \frac{3}{2}\tan B = \frac{3}{2}t$, [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)$.
Standard 11
Mathematics
