Prove 2tan⁻¹(1/5) + tan⁻¹(1/8) = tan⁻¹(4/7)

Prove that \( 2\tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left(\frac{4}{7}\right) \)

Solution:

Let

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

Then,

\[ \tan \theta = \frac{1}{5} \]

Using double angle identity:

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

\[ = \frac{2 \cdot \frac{1}{5}}{1 – \frac{1}{25}} = \frac{2/5}{24/25} = \frac{5}{12} \]

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

Now consider:

\[ \tan^{-1}\left(\frac{5}{12}\right) + \tan^{-1}\left(\frac{1}{8}\right) \]

Using identity:

\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a + b}{1 – ab}\right) \]

\[ = \tan^{-1}\left(\frac{\frac{5}{12} + \frac{1}{8}}{1 – \frac{5}{12}\cdot\frac{1}{8}}\right) \]

\[ = \tan^{-1}\left(\frac{\frac{10 + 3}{24}}{1 – \frac{5}{96}}\right) \]

\[ = \tan^{-1}\left(\frac{13/24}{91/96}\right) = \tan^{-1}\left(\frac{13 \cdot 96}{24 \cdot 91}\right) \]

\[ = \tan^{-1}\left(\frac{52}{91}\right) \]

\[ = \tan^{-1}\left(\frac{4}{7}\right) \]

Hence,

\[ 2\tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left(\frac{4}{7}\right) \]

\[ 2\tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \tan^{-1}\left(\frac{4}{7}\right) \]

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