Prove 2tan⁻¹(3/4) − tan⁻¹(17/31) = π/4

Prove that \( 2\tan^{-1}\left(\frac{3}{4}\right) – \tan^{-1}\left(\frac{17}{31}\right) = \frac{\pi}{4} \)

Solution:

Let

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

Then,

\[ \tan \theta = \frac{3}{4} \]

Using identity:

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

\[ = \frac{2 \cdot \frac{3}{4}}{1 – \left(\frac{3}{4}\right)^2} = \frac{6/4}{1 – 9/16} = \frac{6/4}{7/16} = \frac{24}{7} \]

\[ \Rightarrow 2\theta = \tan^{-1}\left(\frac{24}{7}\right) \]

Now consider:

\[ \tan^{-1}\left(\frac{24}{7}\right) – \tan^{-1}\left(\frac{17}{31}\right) \]

Using identity:

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

\[ = \tan^{-1}\left(\frac{\frac{24}{7} – \frac{17}{31}}{1 + \frac{24}{7}\cdot\frac{17}{31}}\right) \]

\[ = \tan^{-1}\left(\frac{\frac{744 – 119}{217}}{1 + \frac{408}{217}}\right) \]

\[ = \tan^{-1}\left(\frac{625/217}{625/217}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

Hence,

\[ 2\tan^{-1}\left(\frac{3}{4}\right) – \tan^{-1}\left(\frac{17}{31}\right) = \frac{\pi}{4} \]

\[ 2\tan^{-1}\left(\frac{3}{4}\right) – \tan^{-1}\left(\frac{17}{31}\right) = \frac{\pi}{4} \]

Spread the love