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sorry to distrub you, but i was surfing the net for the triple angle of tangent trig function but could not find it so i decided to use infomation i knew to solve it. i would like to know if what i have done is correct, and if it can be simplified further, thanks. What i have done is as follows:

[tex]\

\begin{array}{c}

\tan 3A = \tan \left( {2A + A} \right) \\

= \frac{{\tan 2A + \tan A}}{{1 - \tan 2A\tan A}} \\

= \frac{{\left( {\frac{{2\tan A}}{{1 - \tan ^2 A}}} \right) + \tan A}}{{1 - \left( {\frac{{2\tan A}}{{1 - \tan ^2 A}}} \right) \cdot \tan A}} \\

= \frac{{\frac{{2\tan A + \left( {\tan A\left( {1 - \tan ^2 A} \right)} \right)}}{{1 - \tan ^2 A}}}}{{\frac{{\left( {1 - \tan ^2 A} \right) - 2\tan ^2 A}}{{1 - \tan ^2 A}}}} \\

= \frac{{2\tan A + \left( {\tan A\left( {1 - \tan ^2 A} \right)} \right)}}{{\left( {1 - \tan ^2 A} \right) - 2\tan ^2 A}} \\

= \frac{{2\tan A + \tan A - \tan ^3 A}}{{1 - \tan A - 2\tan ^2 A}} \\

= \frac{{3\tan A - \tan ^3 A}}{{1 - \tan A - 2\tan ^2 A}} \\

\end{array}

\[/tex]

thank you,

Pavadrin

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# Triple angle of tangent

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