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Prove the following.
If tan⁻¹((x-1)/(x+1)) + tan⁻¹((2x-1)/(2x+1)) = tan⁻¹((23)/(36)) then prove that 24x² - 23x - 12 = 0
If $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$ then prove that $24x^2 - 23x - 12 = 0$
Given: $\tan^{-1}\left(\frac{x-1}{x+1}\right) + \tan^{-1}\left(\frac{2x-1}{2x+1}\right) = \tan^{-1}\left(\frac{23}{36}\right)$
To show: $24x^2 - 23x - 12 = 0$
Answer
Answer
Using $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}$ with $A = \frac{x-1}{x+1}$, $B = \frac{2x-1}{2x+1}$:
Numerator: $(x-1)(2x+1) + (2x-1)(x+1) = (2x^2 - x - 1) + (2x^2 + x - 1) = 4x^2 - 2$
Denominator: $(x+1)(2x+1) - (x-1)(2x-1) = (2x^2 + 3x + 1) - (2x^2 - 3x + 1) = 6x$
$\frac{4x^2-2}{6x} = \frac{23}{36}$
$36(4x^2-2) = 138x \Rightarrow 144x^2 - 138x - 72 = 0$
Dividing by 6: $24x^2 - 23x - 12 = 0$. Hence proved.
- Step 1: Use $\tan^{-1}A + \tan^{-1}B = \tan^{-1}\frac{A+B}{1-AB}$ with $A = \frac{x-1}{x+1}$ and $B = \frac{2x-1}{2x+1}$.
- Step 2: Numerator $A+B$ over the common denominator: $(x-1)(2x+1) + (2x-1)(x+1) = (2x^2 - x - 1) + (2x^2 + x - 1) = 4x^2 - 2$.
- Step 3: Denominator $1-AB$ over the common denominator: $(x+1)(2x+1) - (x-1)(2x-1) = (2x^2 + 3x + 1) - (2x^2 - 3x + 1) = 6x$.
- Step 4: So $\tan^{-1}\frac{4x^2-2}{6x} = \tan^{-1}\frac{23}{36}$, giving $\frac{4x^2-2}{6x} = \frac{23}{36}$.
- Step 5: Cross-multiplying: $36(4x^2-2) = 23 \times 6x$, i.e. $144x^2 - 72 = 138x$.
- Step 6: Rearranging: $144x^2 - 138x - 72 = 0$.
- Step 7: Dividing by 6: $24x^2 - 23x - 12 = 0$. Hence proved.