Aspire Faculty ID #16429 · Topic: NIMCET 2009 · Just now
NIMCET 2009

If $\tan^{-1}(2x) + \tan^{-1}(3x) = \dfrac{\pi}{4}$, then $x$ is:

Solution

Use formula: $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\dfrac{a+b}{1-ab}\right)$ 
So: $\tan^{-1}\left(\dfrac{2x+3x}{1 - 6x^2}\right) = \dfrac{\pi}{4}$ 
Thus: $\dfrac{5x}{1 - 6x^2} = 1$ 
Solve: $5x = 1 - 6x^2$ 
$6x^2 + 5x - 1 = 0$ 
 Quadratic gives roots: $x = \dfrac{1}{3}$ or $x = -\dfrac{1}{2}$

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