Prove 2tan⁻¹(1/2) + tan⁻¹(1/7) = tan⁻¹(31/17)

Prove that \( 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right) \)

Solution:

Let

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

Then,

\[ \tan \theta = \frac{1}{2} \]

Using double angle identity:

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

\[ = \frac{2 \cdot \frac{1}{2}}{1 – \frac{1}{4}} = \frac{1}{3/4} = \frac{4}{3} \]

\[ \Rightarrow 2\theta = \tan^{-1}\left(\frac{4}{3}\right) \]

Now consider:

\[ \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) \]

Using identity:

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

\[ = \tan^{-1}\left(\frac{\frac{4}{3} + \frac{1}{7}}{1 – \frac{4}{3}\cdot\frac{1}{7}}\right) \]

\[ = \tan^{-1}\left(\frac{\frac{28 + 3}{21}}{1 – \frac{4}{21}}\right) \]

\[ = \tan^{-1}\left(\frac{31/21}{17/21}\right) = \tan^{-1}\left(\frac{31}{17}\right) \]

Hence,

\[ 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right) \]

\[ 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{31}{17}\right) \]

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