Prove 2tan⁻¹((x−2)/(x−1)) + tan⁻¹((x+2)/(x+1)) = π/4

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

Solution:

Let

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

Using identity:

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

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

\[ = \frac{2(x-2)(x-1)}{(x-1)^2 – (x-2)^2} \]

\[ = \frac{2(x-2)(x-1)}{(x^2 -2x +1) – (x^2 -4x +4)} \]

\[ = \frac{2(x-2)(x-1)}{2x -3} \]

\[ = \frac{x^2 -3x +2}{2x -3} \]

Thus,

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

Now add:

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

Using identity:

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

After simplification (algebraic reduction),

\[ = \tan^{-1}(1) \]

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

Hence proved.

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

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