Prove tan⁻¹(1/7) + 2tan⁻¹(1/3) = π/4

Prove that \( \tan^{-1}\left(\frac{1}{7}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4} \)

Solution:

Let

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

Then,

\[ \tan \theta = \frac{1}{3} \]

Using double angle identity:

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

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

\[ = \frac{2/3}{1 – 1/9} = \frac{2/3}{8/9} = \frac{2}{3} \times \frac{9}{8} = \frac{3}{4} \]

Thus,

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

Now consider:

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

Using identity:

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

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

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

\[ = \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1) \]

\[ = \frac{\pi}{4} \]

Hence,

\[ \tan^{-1}\left(\frac{1}{7}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4} \]

\[ \tan^{-1}\left(\frac{1}{7}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{4} \]

Spread the love