Prove tan⁻¹(2/3) = ½tan⁻¹(12/5)

Prove that \( \tan^{-1}\left(\frac{2}{3}\right) = \frac{1}{2}\tan^{-1}\left(\frac{12}{5}\right) \)

Solution:

Let

\[ \theta = \tan^{-1}\left(\frac{2}{3}\right) \]

Then,

\[ \tan \theta = \frac{2}{3} \]

Using identity:

\[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \]

\[ = \frac{2 \cdot \frac{2}{3}}{1 – \left(\frac{2}{3}\right)^2} \]

\[ = \frac{4/3}{1 – 4/9} \]

\[ = \frac{4/3}{5/9} \]

\[ = \frac{4}{3} \times \frac{9}{5} \]

\[ = \frac{12}{5} \]

Thus,

\[ \tan(2\theta) = \frac{12}{5} \]

Taking inverse tangent:

\[ 2\theta = \tan^{-1}\left(\frac{12}{5}\right) \]

Hence,

\[ \theta = \frac{1}{2}\tan^{-1}\left(\frac{12}{5}\right) \]

But \(\theta = \tan^{-1}(2/3)\), therefore:

\[ \tan^{-1}\left(\frac{2}{3}\right) = \frac{1}{2}\tan^{-1}\left(\frac{12}{5}\right) \]

\[ \tan^{-1}\left(\frac{2}{3}\right) = \frac{1}{2}\tan^{-1}\left(\frac{12}{5}\right) \]

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