MCQ
જો $\tan A = \frac{1}{2},$ તો $\tan 3A = $
- A$\frac{9}{2}$
- ✓$\frac{{11}}{2}$
- C$\frac{7}{2}$
- D$ - \frac{1}{2}$
$ \Rightarrow \tan 3A = \frac{{3\tan A - {{\tan }^3}A}}{{1 - {{3+an }^2}A}} $
$= \frac{{3.\frac{1}{2} - \frac{1}{8}}}{{1 - 3.\frac{1}{4}}} $
$= \frac{{12 - 1}}{2} = \frac{{11}}{2}$.
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